Class 10 Exercise 1.2 Solutions — Complex Numbers

Concepts

Rule

Multiplication by a real number

If $z=x+iy$ and $k$ is real, then $kz=k(x+iy)=kx+i(ky)$ Multiply both coordinates by the same real number.

For $z\neq0$ and $k>0$ , the direction of the arrow from the origin stays the same and its length is multiplied by $k$ . For $k<0$ , the arrow points in the opposite direction and its length is multiplied by $|k|$ . If $k=0$ or $z=0$ , the result is the origin.

Examples

$2(3-2i)=6-4i$ .
$\tfrac12(4+6i)=2+3i$ .
$-1(3-2i)=-3+2i$ .

Rule

Adding and subtracting complex numbers

Let $z_{1}=x_{1}+iy_{1}$ and $z_{2}=x_{2}+iy_{2}$ , with real coordinates. Addition and subtraction act on corresponding parts: $\begin{aligned}z_{1}+z_{2}&=(x_{1}+x_{2})+i(y_{1}+y_{2}), \\ z_{1}-z_{2}&=(x_{1}-x_{2})+i(y_{1}-y_{2})\end{aligned}$

In words: add the real parts together, and add the imaginary parts together. Real parts and imaginary parts never mix.

This is the same as collecting like terms in algebra. The numbers without $i$ are one kind of term, and the numbers with $i$ are another.

Examples

$(2+5i)+(3-i)=(2+3)+(5-1)i=5+4i$
$(9-2i)-(7-3i)=(9-7)+(-2+3)i=2+i$
$(4+0i)+(0+6i)=4+6i$

Rule

Multiplying complex numbers

Let $z_{1}=x_{1}+iy_{1}$ and $z_{2}=x_{2}+iy_{2}$ . All four coordinates are real. Then $z_{1}z_{2}=(x_{1}x_{2}-y_{1}y_{2})+i(x_{1}y_{2}+y_{1}x_{2})$

The formula can be derived: expand the brackets term by term, replace $i^{2}$ by $-1$ , and collect corresponding terms.

The minus sign in $x_{1}x_{2}-y_{1}y_{2}$ comes from $i^{2}=-1$ . It is not an extra rule to learn.

Examples

$(3-4i)(5-6i)=15-18i-20i+24i^{2}=-9-38i$
$(2+3i)(4+7i)=(8-21)+(14+12)i=-13+26i$
$(4-i)(4+i)=16-i^{2}=16+1=17$

Rule

Dividing complex numbers

Let $z_{1}=x_{1}+iy_{1}$ and $z_{2}=x_{2}+iy_{2}$ , with real coordinates and $z_{2}\neq0$ . Multiply the numerator and denominator by $\bar z_{2}=x_{2}-iy_{2}$ : $\frac{z_{1}}{z_{2}}=\frac{(x_{1}+iy_{1})(x_{2}-iy_{2})}{x_{2}^{2}+y_{2}^{2}}$ Expanding the numerator gives $\frac{z_{1}}{z_{2}}=\frac{x_{1}x_{2}+y_{1}y_{2}}{x_{2}^{2}+y_{2}^{2}}+i\frac{y_{1}x_{2}-x_{1}y_{2}}{x_{2}^{2}+y_{2}^{2}}$

The method is what matters: multiply the top and the bottom by the conjugate of the bottom. The conjugate of $x+iy$ is $x-iy$ .

This works because $(x+iy)(x-iy)=x^{2}+y^{2}$ , which is an ordinary real number. The $i$ disappears from the denominator.

Since $z_{2}\neq0$ , its conjugate is also nonzero and $x_{2}^{2}+y_{2}^{2}>0$ . Thus the fraction used to multiply numerator and denominator equals $1$ .

Examples

$\frac{3+4i}{5-7i}=\frac{3+4i}{5-7i}\times\frac{5+7i}{5+7i}=\frac{-13+41i}{74}$
$(5+2i)\div(6-3i)=\frac{5+2i}{6-3i}\times\frac{6+3i}{6+3i}=\frac{24+27i}{45}$

Definition

Additive inverse

The additive inverse of $z$ is the number $-z$ that you add to $z$ to get $0$ : $z+(-z)=0$ For $z=x+iy$ , the additive inverse is $-z=-x-iy$ .

Change the sign of both parts, not just one. This is a common mistake in the whole exercise.

The number $0=0+0i$ is called the additive identity, because $z+0=z$ for every complex number $z$ .

Examples

The additive inverse of $3+2i$ is $-3-2i$ .
The additive inverse of $4-3i$ is $-4+3i$ .
The additive inverse of $-6i$ is $6i$ .

Definition

Multiplicative inverse

For $z\neq 0$ , the multiplicative inverse of $z$ is the number $z^{-1}$ that you multiply $z$ by to get $1$ : $z\,z^{-1}=1, \qquad z^{-1}=\frac{1}{z}$

To find it, write $\frac{1}{z}$ and then clear the $i$ from the denominator with the conjugate, exactly as in division.

For $z=x+iy$ this gives $z^{-1}=\frac{x}{x^{2}+y^{2}}-i\,\frac{y}{x^{2}+y^{2}}$ .

The number $1=1+0i$ is called the multiplicative identity, because $z\times 1=z$ for every complex number $z$ .

Only $0$ has no multiplicative inverse, because you cannot divide by $0$ .

Examples

The multiplicative inverse of $4-3i$ is $\frac{4}{25}+\frac{3}{25}i$ .
The multiplicative inverse of $i$ is $-i$ , because $i\times(-i)=-i^{2}=1$ .

Rule

Properties of complex numbers

For all complex numbers $z_{1}$ , $z_{2}$ and $z_{3}$ , addition and multiplication both satisfy closure, the commutative property and the associative property, each has an identity, and multiplication is distributive over addition.

Closure: $z_{1}+z_{2}$ and $z_{1}z_{2}$ are again complex numbers.

Commutative: $z_{1}+z_{2}=z_{2}+z_{1}$ and $z_{1}z_{2}=z_{2}z_{1}$ . Order does not matter.

Associative: $(z_{1}+z_{2})+z_{3}=z_{1}+(z_{2}+z_{3})$ and $(z_{1}z_{2})z_{3}=z_{1}(z_{2}z_{3})$ . Grouping does not matter.

Identities: $z+0=z$ and $z\times 1=z$ .

Distributive: $z_{1}(z_{2}+z_{3})=z_{1}z_{2}+z_{1}z_{3}$ .

Examples

Commutative: $(2+i)(3-i)=7+i$ and $(3-i)(2+i)=7+i$ .
Identity: $(5-2i)+(0+0i)=5-2i$ .

Deep understanding

Addition and subtraction on the complex plane

The number $x+iy$ corresponds to the point $(x,y)$ and to the arrow from the origin to that point. To add two numbers geometrically, place the tail of the second arrow at the head of the first, without changing its direction or length.

The final endpoint has coordinates $(x_{1}+x_{2},y_{1}+y_{2})$ . This is the same coordinate addition used in the algebraic rule. If the arrows are parallel, the usual parallelogram picture becomes a line segment, but the coordinate rule still works.

To subtract $z_{2}$ , add $-z_{2}$ instead. Negating a number reverses both coordinates, so the arrow is turned through $180^\circ$ about the origin. For example, $-(2-3i)=-2+3i$ .

Why multiplying is just ordinary algebra

Multiply two brackets the way you always have. Each term in the first multiplies each term in the second, giving four products.

Take $z_{1}=x_{1}+iy_{1}$ and $z_{2}=x_{2}+iy_{2}$ . $\begin{aligned}z_{1}z_{2} &= (x_{1}+iy_{1})(x_{2}+iy_{2}) \\ &= x_{1}x_{2}+ix_{1}y_{2}+iy_{1}x_{2}+i^{2}y_{1}y_{2}\end{aligned}$

Only the last term is new. It carries $i^{2}$ , and $i^{2}=-1$ , so that term becomes $-y_{1}y_{2}$ , a real number with no $i$ on it.

Now collect. The terms without $i$ are $x_{1}x_{2}$ and $-y_{1}y_{2}$ . The terms with $i$ are $ix_{1}y_{2}$ and $iy_{1}x_{2}$ . $\begin{aligned}z_{1}z_{2} &= (x_{1}x_{2}-y_{1}y_{2})+i(x_{1}y_{2}+y_{1}x_{2})\end{aligned}$

That is the whole formula. You can reconstruct the formula by expanding the brackets and using $i^{2}=-1$ .

Why the conjugate clears the denominator

The fraction $\frac{3+4i}{5-7i}$ is already a valid complex number. The task is to rewrite it in rectangular form, with a real denominator, so its real and imaginary parts are explicit.

The trick is to multiply the top and the bottom by the conjugate of the bottom. The conjugate of $5-7i$ is $5+7i$ : same real part, opposite sign on the imaginary part.

Look at what the bottom becomes. It is of the form $(a-b)(a+b)$ , which is always $a^{2}-b^{2}$ . $\begin{aligned}(5-7i)(5+7i) &= 5^{2}-(7i)^{2} \\ &= 25-49i^{2} \\ &= 25-49(-1) \\ &= 25+49 = 74\end{aligned}$

Read the third line carefully. Because $i^{2}=-1$ , the subtraction turns into an addition, and the answer is a plain positive number with no $i$ in it at all. That is exactly what we wanted.

In general $(x+iy)(x-iy)=x^{2}+y^{2}$ , which is always real and never negative.

You are allowed to multiply top and bottom because $\frac{5+7i}{5+7i}=1$ , and multiplying by $1$ changes how a number looks but not its value.

The two inverses are different things

Students mix these up, so hold them apart.

The additive inverse undoes addition. You add it to $z$ and land on $0$ , the additive identity. For $z=x+iy$ it is $-x-iy$ . Both signs change, and no division is needed.

The multiplicative inverse undoes multiplication. You multiply $z$ by it and land on $1$ , the multiplicative identity. It is $\frac{1}{z}$ for $z\neq0$ . The conjugate method rewrites this reciprocal in rectangular form.

A quick check for the multiplicative inverse: multiply your answer by $z$ and see whether you get $1$ . $\begin{aligned}(4-3i)\left(\frac{4}{25}+\frac{3}{25}i\right) &= \frac{(4-3i)(4+3i)}{25} \\ &= \frac{16+9}{25} \\ &= 1 \quad \checkmark\end{aligned}$

Every complex number has an additive inverse. Every complex number except $0$ has a multiplicative inverse, because you cannot divide by $0$ .

Which algebraic rules still apply?

Complex addition and multiplication preserve the commutative, associative, and distributive laws. That is why expanding brackets, collecting terms, and solving linear equations still work.

Not every rule for real numbers transfers unchanged. Complex numbers do not have a compatible order like the real number line, and the square-root product rule needs care. Always keep the hypotheses of the rule being used.

Examples

Example 3

Add $3+4i$ and $5-2i$ .

Solution

Compact solution

There is a plus before the second bracket, so neither sign changes when we remove the brackets.
$(3+4i)+(5-2i)=3+4i+5-2i$
Collect the real terms together and the terms carrying $i$ together.
$(3+4i)+(5-2i)=(3+5)+(4-2)i$
The real terms total $8$ . The coefficient of $i$ is $2$ , so write the result in rectangular form.
$(3+4i)+(5-2i)=8+2i$

Answer

8+2i

Compact solution

Add the real coefficients and the imaginary coefficients separately.

\begin{aligned}(3+4i)+(5-2i) &= 3+4i+5-2i \\ &= (3+5)+(4-2)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 8+2i\end{aligned}

Example 4

Simplify:

(i)

(8+2i)-(5-6i)

Solution

Compact solution

The minus before the second bracket changes every sign inside: $-(5-6i)=-5+6i$ .
$(8+2i)-(5-6i)=8+2i-5+6i$
Collect real terms together and imaginary terms together.
$(8+2i)-(5-6i)=(8-5)+(2+6)i$
The real part is $3$ and the imaginary part is $8$ .
$(8+2i)-(5-6i)=3+8i$

Answer

3+8i

Compact solution

Remove the second bracket by changing both signs, then collect like terms.

\begin{aligned}(8+2i)-(5-6i) &= 8+2i-5+6i \\ &= (8-5)+(2+6)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 3+8i\end{aligned}

(ii)

(4-3i)-(2-5i)

Solution

Compact solution

The minus before the second bracket changes both signs. In particular, subtracting $-5i$ adds $5i$ .
$(4-3i)-(2-5i)=4-3i-2+5i$
Collect real terms together and imaginary terms together.
$(4-3i)-(2-5i)=(4-2)+(-3+5)i$
For $-3+5$ , start at $-3$ and move $5$ to the right to reach $2$ . The real part is also $2$ .
$(4-3i)-(2-5i)=2+2i$

Answer

2+2i

Compact solution

Remove the second bracket by changing both signs, then collect like terms.

\begin{aligned}(4-3i)-(2-5i) &= 4-3i-2+5i \\ &= (4-2)+(-3+5)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 2+2i\end{aligned}

Example 5

Multiply the following:

(i)

(3-4i)(5-6i)

Solution

Compact solution

Multiply the brackets out term by term. Each term in the first bracket multiplies each term in the second, which gives four products.
$(3-4i)(5-6i) = 3(5) + 3(-6i) + (-4i)(5) + (-4i)(-6i)$
Work out the first three products. In the second and third, plus times minus gives minus.
$3(5) = 15, \quad 3(-6i) = -18i, \quad (-4i)(5) = -20i$
Now the fourth. Minus times minus gives plus, and $4\times 6=24$ .
$(-4i)(-6i) = 24i^{2}$
Put the four products together.
$(3-4i)(5-6i) = 15-18i-20i+24i^{2}$
Replace $i^{2}$ by $-1$ in the last term.
$24i^{2} = 24(-1) = -24$
Now collect. Put the real numbers together and the $i$ terms together.
$(3-4i)(5-6i) = (15-24)+(-18-20)i$
Taking $24$ from $15$ carries you $9$ below $0$ , and $-18-20=-38$ .
$(3-4i)(5-6i) = -9-38i$

Answer

-9-38i

Compact solution

Multiply each term in the first bracket by each term in the second.

\begin{aligned}(3-4i)(5-6i) &= 3(5)+3(-6i)+(-4i)(5)+(-4i)(-6i) \\ &= 15-18i-20i+24i^2 \\ &= 15-18i-20i+24(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 15-18i-20i-24 \\ &= (15-24)+(-18-20)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= -9-38i\end{aligned}

(ii)If

z_{1}=2+3i

and

z_{2}=4+7i

, find

z_{1}z_{2}

Solution

Compact solution

Write down the four numbers first, matching them to the form $x+iy$ .
$x_{1}=2, \quad y_{1}=3, \quad x_{2}=4, \quad y_{2}=7$
You may use the formula, or simply multiply the brackets out. Here is the formula.
$z_{1}z_{2} = (x_{1}x_{2}-y_{1}y_{2})+i(x_{1}y_{2}+y_{1}x_{2})$
Work out the real part. Note the minus sign in the middle, which comes from $i^{2}=-1$ .
$x_{1}x_{2}-y_{1}y_{2} = (2)(4)-(3)(7) = 8-21$
Taking $21$ from $8$ carries you $13$ below $0$ .
$x_{1}x_{2}-y_{1}y_{2} = -13$
Now the imaginary part.
$x_{1}y_{2}+y_{1}x_{2} = (2)(7)+(3)(4) = 14+12 = 26$
Put the two parts together.
$z_{1}z_{2} = -13+26i$

Answer

-13+26i

Compact solution

Use $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ . Here

\begin{gathered}x_1=2\quad y_1=3 \\ x_2=4\quad y_2=7\end{gathered}

Substitute into the multiplication formula.

\begin{aligned}z_1z_2 &= (x_1x_2-y_1y_2)+i(x_1y_2+y_1x_2) \\ &= (2\cdot4-3\cdot7)+i(2\cdot7+3\cdot4) \\ &= (8-21)+i(14+12) \\ &= -13+26i\end{aligned}

Other forms

Also correctMultiplying the brackets out directly is equally correct and gives the same answer:

(2+3i)(4+7i)=8+14i+12i+21i^{2}=-13+26i

Example 6

Express $\frac{3+4i}{5-7i}$ in the form $x+iy$ .

Solution

Compact solution

To express the result in rectangular form, make the denominator real. Clear it by multiplying the top and the bottom by the conjugate of the bottom. The conjugate of $5-7i$ is $5+7i$ .
$\frac{3+4i}{5-7i} = \frac{3+4i}{5-7i}\times\frac{5+7i}{5+7i}$
This is allowed because $\frac{5+7i}{5+7i}=1$ , and multiplying by $1$ changes how a number looks but not its value.
$\frac{5+7i}{5+7i} = 1$
Work out the numerator first. Each term in the first bracket multiplies each term in the second, giving four products.
$(3+4i)(5+7i) = 3(5)+3(7i)+4i(5)+4i(7i)$
Work out the first three products.
$3(5) = 15, \quad 3(7i) = 21i, \quad 4i(5) = 20i$
Now the fourth. Multiply the numbers, and the two $i$ terms give $i^{2}$ .
$4i(7i) = 28i^{2}$
Put the four products together.
$(3+4i)(5+7i) = 15+21i+20i+28i^{2}$
Replace $i^{2}$ by $-1$ in the last term.
$28i^{2} = 28(-1) = -28$
Collect. The real numbers are $15$ and $-28$ , and the $i$ terms are $21i$ and $20i$ .
$(3+4i)(5+7i) = (15-28)+(21+20)i = -13+41i$
Now the denominator. The two brackets are of the form $(a-b)(a+b)$ , which is always $a^{2}-b^{2}$ .
$(5-7i)(5+7i) = 5^{2}-(7i)^{2}$
Square each part. For the second, square both the $7$ and the $i$ .
$5^{2} = 25 \quad\text{and}\quad (7i)^{2} = 49i^{2} = 49(-1) = -49$
Subtract. Taking away $-49$ is the same as adding $49$ , so the denominator is a real number with no $i$ in it.
$(5-7i)(5+7i) = 25-(-49) = 74$
Put the numerator over the denominator.
$\frac{3+4i}{5-7i} = \frac{-13+41i}{74}$
Finally split it into a real part and an imaginary part, as required.
$\frac{3+4i}{5-7i} = -\frac{13}{74}+\frac{41}{74}i$

Answer

-\dfrac{13}{74}+\dfrac{41}{74}i

Compact solution

Multiply numerator and denominator by the conjugate $5+7i$ .

\begin{aligned}\frac{3+4i}{5-7i} &= \frac{3+4i}{5-7i}\cdot\frac{5+7i}{5+7i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{5+7i}{5+7i}=1$}} \\ &= \frac{(3+4i)(5+7i)}{(5-7i)(5+7i)}\end{aligned}

Expand the numerator first.

\begin{aligned}(3+4i)(5+7i) &= 3(5)+3(7i)+(4i)(5)+(4i)(7i) \\ &= 15+21i+20i+28i^2 \\ &= 15+21i+20i+28(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 15+21i+20i-28 \\ &= (15-28)+(21+20)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= -13+41i\end{aligned}

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(5-7i)(5+7i) &= 5^2-(7i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 25-49i^2 \\ &= 25-49(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 25+49 \\ &= 74\end{aligned}

Substitute the numerator and denominator, then simplify to rectangular form.

\begin{aligned}\frac{3+4i}{5-7i} &= \frac{-13+41i}{74} \\ &= -\frac{13}{74}+\frac{41}{74}i\end{aligned}

Example 7

If $z=-7-4i$ , show that $z+0=z$ .

Solution

Compact solution

Write $0$ in full complex form, so that both numbers are in the shape $x+iy$ and can be added part by part.
$0 = 0+0i$
Now add. Add the real parts together and the imaginary parts together.
$z+0 = (-7-4i)+(0+0i)$
Take the real parts. Adding $0$ changes nothing.
$-7+0 = -7$
Take the imaginary parts. Adding $0$ changes nothing here either.
$-4+0 = -4$
Put the two parts back together, which is the number we started with.
$z+0 = -7-4i = z$

Answer

z+0=z

, so

0

is the additive identity.

Compact solution

Write the additive identity as $0=0+0i$ and add corresponding coefficients.

\begin{aligned}z+0 &= (-7-4i)+(0+0i) \\ &= (-7+0)+(-4+0)i \\ &= -7-4i \\ &= z\end{aligned}

Observation

ObservationThis is why

0

is called the additive identity. It leaves every complex number unchanged, just as

0

does for real numbers.

Example 8

Verify the multiplicative identity for $z=3-2i$ .

Solution

Compact solution

Write $1$ in full complex form.
$1 = 1+0i$
Multiply $z$ by $1$ , taking the brackets out term by term.
$z\times 1 = (3-2i)(1+0i) = 3+0i-2i-0i^{2}$
The two terms carrying $0$ are both $0$ , and so is the $i^{2}$ term.
$z\times 1 = 3-2i = z$
Now multiply the other way round, to check that the order makes no difference.
$1\times z = (1+0i)(3-2i) = 3-2i+0i-0i^{2}$
Again the terms equal to $0$ vanish.
$1\times z = 3-2i = z$
Both products give back $z$ , which is what was to be shown.
$z\times 1 = 1\times z = z$

Answer

z\times 1=1\times z=z

, so

1

is the multiplicative identity.

Compact solution

Write the multiplicative identity as $1=1+0i$ . Multiply in the first order.

\begin{aligned}z\cdot1 &= (3-2i)(1+0i) \\ &= 3+0i-2i-0i^2 \\ &= 3-2i \\ &= z\end{aligned}

Now multiply in the opposite order.

\begin{aligned}1\cdot z &= (1+0i)(3-2i) \\ &= 3-2i+0i-0i^2 \\ &= 3-2i \\ &= z\end{aligned}

Both orders leave $z$ unchanged.

z\cdot1=1\cdot z=z

Example 9

Find the additive inverse of $z=7-10i$ .

Solution

Compact solution

The additive inverse is the number you add to $z$ to get $0$ . Call it $-z$ .
$z+(-z) = 0$
To find it, change the sign of every term in $z$ . The real part $7$ becomes $-7$ .
$7 \to -7$
The imaginary part $-10$ becomes $+10$ . Both signs change, not just one.
$-10 \to +10$
So the additive inverse is this.
$-z = -7+10i$

Answer

-7+10i

Compact solution

Let $z=7-10i$ . The additive inverse changes both signs, so that its sum with the original number is $0$ .

\begin{aligned}-z &= -(7-10i) \\ &= -7+10i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$-(a-b)=-a+b$}}\end{aligned}

Check

Check the answer

\begin{aligned}z+(-z) &= (7-10i)+(-7+10i) \\ &= (7-7)+(-10+10)i \\ &= 0 \quad \checkmark\end{aligned}

Adding the two together gives

0

, so the answer is correct.

Example 10

Find the multiplicative inverse of $z=4-3i$ .

Solution

Compact solution

The multiplicative inverse is the number you multiply $z$ by to get $1$ . It is written $z^{-1}$ and equals $\frac{1}{z}$ .
$z^{-1} = \frac{1}{4-3i}$
The reciprocal is correct; next write it in rectangular form. Multiply top and bottom by the conjugate of the bottom, which is $4+3i$ .
$z^{-1} = \frac{1}{4-3i}\times\frac{4+3i}{4+3i}$
Take the denominator first. It is of the form $(a-b)(a+b)$ , which is always $a^{2}-b^{2}$ .
$(4-3i)(4+3i) = 4^{2}-(3i)^{2}$
Square each part, remembering $i^{2}=-1$ .
$4^{2} = 16 \quad\text{and}\quad (3i)^{2} = 9i^{2} = -9$
Subtract. Taking away $-9$ is the same as adding $9$ .
$(4-3i)(4+3i) = 16+9 = 25$
The numerator is simply $4+3i$ , because the top was $1$ .
$z^{-1} = \frac{4+3i}{25}$
Split it into a real part and an imaginary part.
$z^{-1} = \frac{4}{25}+\frac{3}{25}i$

Answer

\dfrac{4}{25}+\dfrac{3}{25}i

Compact solution

For $z=4-3i$ , find $z^{-1}=1/z$ . Write the reciprocal and use the conjugate $4+3i$ .

\begin{aligned}z^{-1} &= \frac{1}{4-3i} \\ &= \frac{1}{4-3i}\cdot\frac{4+3i}{4+3i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{4+3i}{4+3i}=1$}} \\ &= \frac{4+3i}{(4-3i)(4+3i)}\end{aligned}

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(4-3i)(4+3i) &= 4^2-(3i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 16-9i^2 \\ &= 16-9(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 16+9 \\ &= 25\end{aligned}

Substitute the numerator and denominator, then simplify to rectangular form.

\begin{aligned}z^{-1} &= \frac{4+3i}{25} \\ &= \frac{4}{25}+\frac{3}{25}i\end{aligned}

Check

Check the answer

\begin{aligned}z\,z^{-1} &= (4-3i)\cdot\frac{4+3i}{25} \\ &= \frac{16+9}{25} \\ &= 1 \quad \checkmark\end{aligned}

Multiplying

z

by the answer gives

1

, so the inverse is correct.

Skilled practice

Apply the addition, multiplication, and division rules.

(i)If

z_{1}-z_{2}=4+6i

and

z_{2}=3-2i

, find

z_{1}

Solution

Compact solution

Add $z_{2}$ to both sides to isolate $z_{1}$ .
$z_{1}=4+6i+z_{2}$
Substitute $z_{2}=3-2i$ .
$z_{1}=(4+6i)+(3-2i)$
Add the real parts and the imaginary parts separately.
$z_{1}=(4+3)+(6-2)i$
The sums are $7$ and $4$ , respectively.
$z_{1}=7+4i$

Answer

z_{1}=7+4i

Compact solution

Add $z_2$ to both sides of the given equation.

z_1=4+6i+z_2

Add the real coefficients and the imaginary coefficients separately.

\begin{aligned}z_1 &= (4+6i)+(3-2i) \\ &= 4+6i+3-2i \\ &= (4+3)+(6-2)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 7+4i\end{aligned}

(ii)Simplify

\frac{1+2i}{3-i}(2+i)

Solution

Compact solution

Combine the factors into one fraction and expand the numerator.
$\frac{1+2i}{3-i}(2+i)=\frac{2+i+4i+2i^{2}}{3-i}$
Use $i^{2}=-1$ ; the real terms cancel.
$\frac{1+2i}{3-i}(2+i)=\frac{5i}{3-i}$
Multiply numerator and denominator by $3+i$ , the conjugate of the denominator.
$\frac{5i}{3-i}=\frac{5i(3+i)}{(3-i)(3+i)}$
The numerator is $15i+5i^{2}=-5+15i$ . The denominator is $9-i^{2}=10$ .
$\frac{5i}{3-i}=\frac{-5+15i}{10}$
Divide both numerator terms by $10$ and reduce the fractions.
$\frac{1+2i}{3-i}(2+i)=-\frac12+\frac32i$

Answer

-\dfrac12+\dfrac32i

Compact solution

Multiply numerator and denominator by the conjugate $3+i$ .

\begin{aligned}\frac{1+2i}{3-i} &= \frac{1+2i}{3-i}\cdot\frac{3+i}{3+i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{3+i}{3+i}=1$}} \\ &= \frac{(1+2i)(3+i)}{(3-i)(3+i)}\end{aligned}

Expand the numerator first.

\begin{aligned}(1+2i)(3+i) &= 1(3)+1(i)+(2i)(3)+(2i)(i) \\ &= 3+i+6i+2i^2 \\ &= 3+i+6i+2(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 3+i+6i-2 \\ &= (3-2)+(1+6)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 1+7i\end{aligned}

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(3-i)(3+i) &= 3^2-i^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 9-i^2 \\ &= 9-1(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 9+1 \\ &= 10\end{aligned}

Substitute the numerator and denominator, then simplify to rectangular form.

\begin{aligned}\frac{1+2i}{3-i} &= \frac{1+7i}{10} \\ &= \frac{1}{10}+\frac{7}{10}i\end{aligned}

Multiply this quotient by the remaining factor $2+i$ . Multiply each term in the first bracket by each term in the second.

\begin{aligned}\frac{1+2i}{3-i}(2+i) &= (\dfrac{1}{10}+\dfrac{7}{10}i)(2+i) \\ &= \dfrac{1}{10}(2)+\dfrac{1}{10}(i)+(\dfrac{7}{10}i)(2)+(\dfrac{7}{10}i)(i) \\ &= \dfrac{1}{5}+\dfrac{1}{10}i+\dfrac{7}{5}i+\dfrac{7}{10}i^2 \\ &= \dfrac{1}{5}+\dfrac{1}{10}i+\dfrac{7}{5}i+\dfrac{7}{10}(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= (\dfrac{1}{5}-(\dfrac{7}{10}))+(\dfrac{1}{10}+\dfrac{7}{5})i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= -\dfrac{1}{2}+\dfrac{3}{2}i\end{aligned}

(iii)If

\frac{z}{2+i}=3-i

, find

z

Solution

Compact solution

Since $2+i\neq0$ , multiply both sides by $2+i$ .
$z=(3-i)(2+i)$
Multiply each term in the first factor by each term in the second.
$z=6+3i-2i-i^{2}$
Use $-i^{2}=1$ and $3i-2i=i$ .
$z=6+i+1$
Collect the real terms.
$z=7+i$

Answer

z=7+i

Compact solution

Since $2+i\neq0$ , multiply both sides by $2+i$ .

z=(3-i)(2+i)

Multiply each term in the first bracket by each term in the second.

\begin{aligned}z &= (3-i)(2+i) \\ &= 3(2)+3(i)+(-i)(2)+(-i)(i) \\ &= 6+3i-2i-i^2 \\ &= 6+3i-2i-1(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 6+3i-2i+1 \\ &= (6+1)+(3-2)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 7+i\end{aligned}

Exercise 1.2

Question 1

Simplify and write in the form $a+bi$ :

(i)

(2+5i)+(3-i)

Solution

Compact solution

Adding two complex numbers means adding the real parts together and adding the imaginary parts together. Real parts and imaginary parts never mix. Start by gathering the real parts, which are the pieces with no $i$ on them.
$\text{Real parts: } 2 + 3$
Gather the imaginary parts, which are the numbers multiplying $i$ , without the $i$ itself.
$\text{Imaginary parts: } 5 + (-1)$
Work out each sum separately.
$2+3 = 5, \qquad 5+(-1) = 4$
Put the two answers back together in the form $a+bi$ .
$(2+5i)+(3-i) = 5 + 4i$

Answer

5 + 4i

Compact solution

Add the real coefficients and the imaginary coefficients separately.

\begin{aligned}(2+5i)+(3-i) &= 2+5i+3-i \\ &= (2+3)+(5-1)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 5+4i\end{aligned}

(ii)

(16-3i)+(9+2i)

Solution

Compact solution

Adding two complex numbers means adding the real parts together and adding the imaginary parts together. Real parts and imaginary parts never mix. Start by gathering the real parts, which are the pieces with no $i$ on them.
$\text{Real parts: } 16 + 9$
Gather the imaginary parts, which are the numbers multiplying $i$ , without the $i$ itself.
$\text{Imaginary parts: } -3 + 2$
Work out each sum separately.
$16+9 = 25, \qquad -3+2 = -1$
Put the two answers back together in the form $a+bi$ .
$(16-3i)+(9+2i) = 25 - i$

Answer

25 - i

Compact solution

Add the real coefficients and the imaginary coefficients separately.

\begin{aligned}(16-3i)+(9+2i) &= 16-3i+9+2i \\ &= (16+9)+(-3+2)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 25-i\end{aligned}

(iii)

(9-2i)-(7-3i)

Solution

Compact solution

Subtracting means taking the second bracket away from the first. The minus sign in front of the bracket changes the sign of every term inside it, so each part of the second number is taken away. Start with the real parts.
$\text{Real parts: } 9 - 7$
Gather the imaginary parts.
$\text{Imaginary parts: } -2 - (-3)$
Work out each one separately.
$9-7 = 2, \qquad -2-(-3) = 1$
Put the two answers back together.
$(9-2i)-(7-3i) = 2 + i$

Answer

2 + i

Compact solution

Remove the second bracket by changing both signs, then collect like terms.

\begin{aligned}(9-2i)-(7-3i) &= 9-2i-7+3i \\ &= (9-7)+(-2+3)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 2+i\end{aligned}

(iv)

(11+9i)-(9-7i)

Solution

Compact solution

Subtracting means taking the second bracket away from the first. The minus sign in front of the bracket changes the sign of every term inside it, so each part of the second number is taken away. Start with the real parts.
$\text{Real parts: } 11 - 9$
Gather the imaginary parts.
$\text{Imaginary parts: } 9 - (-7)$
Work out each one separately.
$11-9 = 2, \qquad 9-(-7) = 16$
Put the two answers back together.
$(11+9i)-(9-7i) = 2 + 16i$

Answer

2 + 16i

Compact solution

Remove the second bracket by changing both signs, then collect like terms.

\begin{aligned}(11+9i)-(9-7i) &= 11+9i-9+7i \\ &= (11-9)+(9+7)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 2+16i\end{aligned}

(v)

(3+4i)(2-3i)

Solution

Compact solution

Multiply the brackets out term by term. Each term in the first multiplies each term in the second, giving four products.
$(3+4i)(2-3i) = 3(2)+3(-3i)+4i(2)+4i(-3i)$
Work out the first three products.
$3(2) = 6, \quad 3(-3i) = -9i, \quad 4i(2) = 8i$
Now the fourth. Plus times minus gives minus, and $4\times 3=12$ .
$4i(-3i) = -12i^{2}$
Put the four products together.
$(3+4i)(2-3i) = 6-9i+8i-12i^{2}$
Replace $i^{2}$ by $-1$ . Note that $-12(-1)=+12$ , because minus times minus gives plus.
$-12i^{2} = -12(-1) = 12$
Collect. The real numbers are $6$ and $12$ , and the $i$ terms are $-9i$ and $8i$ .
$(3+4i)(2-3i) = (6+12)+(-9+8)i$
And $6+12=18$ , while $-9+8=-1$ .
$(3+4i)(2-3i) = 18-i$

Answer

18-i

Compact solution

Multiply each term in the first bracket by each term in the second.

\begin{aligned}(3+4i)(2-3i) &= 3(2)+3(-3i)+(4i)(2)+(4i)(-3i) \\ &= 6-9i+8i-12i^2 \\ &= 6-9i+8i-12(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 6-9i+8i+12 \\ &= (6+12)+(-9+8)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 18-i\end{aligned}

Common mistake

Watch outA common mistake here is writing

-12i^{2}=-12

. Two minus signs are at work: the one already in front of the

12

, and the one from

i^{2}=-1

. They cancel, so the term is

+12

(vi)

(5-2i)(3-4i)

Solution

Compact solution

Multiply the brackets out term by term.
$(5-2i)(3-4i) = 5(3)+5(-4i)+(-2i)(3)+(-2i)(-4i)$
Work out the first three products.
$5(3) = 15, \quad 5(-4i) = -20i, \quad (-2i)(3) = -6i$
Now the fourth. Minus times minus gives plus, and $2\times 4=8$ .
$(-2i)(-4i) = 8i^{2}$
Put the four products together.
$(5-2i)(3-4i) = 15-20i-6i+8i^{2}$
Replace $i^{2}$ by $-1$ .
$8i^{2} = 8(-1) = -8$
Collect the real numbers and the $i$ terms.
$(5-2i)(3-4i) = (15-8)+(-20-6)i$
And $15-8=7$ , while $-20-6=-26$ , because going down $6$ more from $-20$ reaches $-26$ .
$(5-2i)(3-4i) = 7-26i$

Answer

7-26i

Compact solution

Multiply each term in the first bracket by each term in the second.

\begin{aligned}(5-2i)(3-4i) &= 5(3)+5(-4i)+(-2i)(3)+(-2i)(-4i) \\ &= 15-20i-6i+8i^2 \\ &= 15-20i-6i+8(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 15-20i-6i-8 \\ &= (15-8)+(-20-6)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 7-26i\end{aligned}

(vii)

(3-5i)\div(2-4i)

Solution

Compact solution

Division is written as a fraction.
$(3-5i)\div(2-4i) = \frac{3-5i}{2-4i}$
To express the result in rectangular form, make the denominator real. Multiply the top and the bottom by the conjugate of the bottom. The conjugate of $2-4i$ is $2+4i$ .
$\frac{3-5i}{2-4i} = \frac{3-5i}{2-4i}\times\frac{2+4i}{2+4i}$
This is allowed because $\frac{2+4i}{2+4i}=1$ , and multiplying by $1$ changes how a number looks but not its value.
$\frac{2+4i}{2+4i} = 1$
Take the denominator first. It is of the form $(a-b)(a+b)$ , which is always $a^{2}-b^{2}$ .
$(2-4i)(2+4i) = 2^{2}-(4i)^{2}$
Square each part, remembering $i^{2}=-1$ .
$2^{2} = 4 \quad\text{and}\quad (4i)^{2} = 16i^{2} = -16$
Subtract. Taking away $-16$ is the same as adding $16$ .
$(2-4i)(2+4i) = 4+16 = 20$
Now the numerator, multiplying out term by term.
$(3-5i)(2+4i) = 6+12i-10i-20i^{2}$
Replace $i^{2}$ by $-1$ . Note $-20(-1)=+20$ .
$-20i^{2} = 20$
Collect the numerator.
$(3-5i)(2+4i) = (6+20)+(12-10)i = 26+2i$
Put the numerator over the denominator.
$\frac{3-5i}{2-4i} = \frac{26+2i}{20}$
Every number here divides by $2$ , so cancel the common factor $2$ from the top and the bottom.
$\frac{26+2i}{20} = \frac{2(13+i)}{2(10)} = \frac{13+i}{10}$
Finally split it into a real part and an imaginary part.
$\frac{3-5i}{2-4i} = \frac{13}{10}+\frac{1}{10}i$

Answer

\dfrac{13}{10}+\dfrac{1}{10}i

Compact solution

Multiply numerator and denominator by the conjugate $2+4i$ .

\begin{aligned}\frac{3-5i}{2-4i} &= \frac{3-5i}{2-4i}\cdot\frac{2+4i}{2+4i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{2+4i}{2+4i}=1$}} \\ &= \frac{(3-5i)(2+4i)}{(2-4i)(2+4i)}\end{aligned}

Expand the numerator first.

\begin{aligned}(3-5i)(2+4i) &= 3(2)+3(4i)+(-5i)(2)+(-5i)(4i) \\ &= 6+12i-10i-20i^2 \\ &= 6+12i-10i-20(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 6+12i-10i+20 \\ &= (6+20)+(12-10)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 26+2i\end{aligned}

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(2-4i)(2+4i) &= 2^2-(4i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 4-16i^2 \\ &= 4-16(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 4+16 \\ &= 20\end{aligned}

Substitute the numerator and denominator, then simplify to rectangular form.

\begin{aligned}\frac{3-5i}{2-4i} &= \frac{26+2i}{20} \\ &= \frac{2(13+i)}{2(10)} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{cancel the common factor }2$}} \\ &= \frac{13+i}{10} \\ &= \frac{13}{10}+\frac{1}{10}i\end{aligned}

(viii)

(5+2i)\div(6-3i)

Solution

Compact solution

Division is written as a fraction.
$(5+2i)\div(6-3i) = \frac{5+2i}{6-3i}$
Clear the $i$ from the denominator by multiplying the top and the bottom by the conjugate of the bottom, which is $6+3i$ .
$\frac{5+2i}{6-3i} = \frac{5+2i}{6-3i}\times\frac{6+3i}{6+3i}$
Take the denominator first, using $(a-b)(a+b)=a^{2}-b^{2}$ .
$(6-3i)(6+3i) = 6^{2}-(3i)^{2}$
Square each part, remembering $i^{2}=-1$ .
$6^{2} = 36 \quad\text{and}\quad (3i)^{2} = 9i^{2} = -9$
Subtract. Taking away $-9$ is the same as adding $9$ .
$(6-3i)(6+3i) = 36+9 = 45$
Now the numerator, term by term.
$(5+2i)(6+3i) = 30+15i+12i+6i^{2}$
Replace $i^{2}$ by $-1$ .
$6i^{2} = -6$
Collect the numerator.
$(5+2i)(6+3i) = (30-6)+(15+12)i = 24+27i$
Put the numerator over the denominator.
$\frac{5+2i}{6-3i} = \frac{24+27i}{45}$
All three numbers divide by $3$ , so cancel the common factor $3$ .
$\frac{24+27i}{45} = \frac{3(8+9i)}{3(15)} = \frac{8+9i}{15}$
Split into a real part and an imaginary part.
$\frac{5+2i}{6-3i} = \frac{8}{15}+\frac{9}{15}i$
The imaginary part can still be reduced, because $9$ and $15$ both divide by $3$ .
$\frac{9}{15} = \frac{3}{5}$
So the finished answer is this.
$\frac{5+2i}{6-3i} = \frac{8}{15}+\frac{3}{5}i$

Answer

\dfrac{8}{15}+\dfrac{3}{5}i

Compact solution

Multiply numerator and denominator by the conjugate $6+3i$ .

\begin{aligned}\frac{5+2i}{6-3i} &= \frac{5+2i}{6-3i}\cdot\frac{6+3i}{6+3i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{6+3i}{6+3i}=1$}} \\ &= \frac{(5+2i)(6+3i)}{(6-3i)(6+3i)}\end{aligned}

Expand the numerator first.

\begin{aligned}(5+2i)(6+3i) &= 5(6)+5(3i)+(2i)(6)+(2i)(3i) \\ &= 30+15i+12i+6i^2 \\ &= 30+15i+12i+6(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 30+15i+12i-6 \\ &= (30-6)+(15+12)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 24+27i\end{aligned}

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(6-3i)(6+3i) &= 6^2-(3i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 36-9i^2 \\ &= 36-9(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 36+9 \\ &= 45\end{aligned}

Substitute the numerator and denominator, then simplify to rectangular form.

\begin{aligned}\frac{5+2i}{6-3i} &= \frac{24+27i}{45} \\ &= \frac{3(8+9i)}{3(15)} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{cancel the common factor }3$}} \\ &= \frac{8+9i}{15} \\ &= \frac{8}{15}+\frac{9}{15}i \\ &= \frac{8}{15}+\frac{3}{5}i\end{aligned}

Common mistake

Watch outReduce

\frac9{15}

by the common factor

3

to get

\frac35

. The earlier fraction has the same value; reducing it gives the requested simplified form.

Question 2

Write the additive inverse of each complex number:

(i)

3+2i

Solution

Compact solution

The additive inverse of a number is what you add to it to get $0$ . For $z$ it is written $-z$ .
$z = 3+2i, \qquad z+(-z) = 0$
To find $-z$ , put a minus in front of the whole number. A minus in front of a bracket changes the sign of every term inside.
$-z = -\left(3+2i\right)$
Change the sign of the real part.
$3 \to -3$
Change the sign of the imaginary part. Both signs change, not just one.
$+2 \to -2$
So the additive inverse is this.
$-z = -3-2i$

Answer

-3-2i

Compact solution

Let $z=3+2i$ . The additive inverse changes both signs, so that its sum with the original number is $0$ .

\begin{aligned}-z &= -(3+2i) \\ &= -3-2i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$-(a+b)=-a-b$}}\end{aligned}

(ii)

4-3i

Solution

Compact solution

The additive inverse of a number is what you add to it to get $0$ . For $z$ it is written $-z$ .
$z = 4-3i, \qquad z+(-z) = 0$
To find $-z$ , put a minus in front of the whole number. A minus in front of a bracket changes the sign of every term inside.
$-z = -\left(4-3i\right)$
Change the sign of the real part.
$4 \to -4$
Change the sign of the imaginary part. Both signs change, not just one.
$-3 \to +3$
So the additive inverse is this.
$-z = -4+3i$

Answer

-4+3i

Compact solution

Let $z=4-3i$ . The additive inverse changes both signs, so that its sum with the original number is $0$ .

\begin{aligned}-z &= -(4-3i) \\ &= -4+3i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$-(a-b)=-a+b$}}\end{aligned}

(iii)

5-7i

Solution

Compact solution

The additive inverse of a number is what you add to it to get $0$ . For $z$ it is written $-z$ .
$z = 5-7i, \qquad z+(-z) = 0$
To find $-z$ , put a minus in front of the whole number. A minus in front of a bracket changes the sign of every term inside.
$-z = -\left(5-7i\right)$
Change the sign of the real part.
$5 \to -5$
Change the sign of the imaginary part. Both signs change, not just one.
$-7 \to +7$
So the additive inverse is this.
$-z = -5+7i$

Answer

-5+7i

Compact solution

Let $z=5-7i$ . The additive inverse changes both signs, so that its sum with the original number is $0$ .

\begin{aligned}-z &= -(5-7i) \\ &= -5+7i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$-(a-b)=-a+b$}}\end{aligned}

(iv)

-\frac{2}{3}+\frac{5}{4}i

Solution

Compact solution

The additive inverse of a number is what you add to it to get $0$ . For $z$ it is written $-z$ .
$z = -\dfrac{2}{3}+\dfrac{5}{4}i, \qquad z+(-z) = 0$
To find $-z$ , put a minus in front of the whole number. A minus in front of a bracket changes the sign of every term inside.
$-z = -\left(-\dfrac{2}{3}+\dfrac{5}{4}i\right)$
Change the sign of the real part.
$-\frac{2}{3} \to +\frac{2}{3}$
Change the sign of the imaginary part. Both signs change, not just one.
$+\frac{5}{4} \to -\frac{5}{4}$
So the additive inverse is this.
$-z = \dfrac{2}{3}-\dfrac{5}{4}i$

Answer

\dfrac{2}{3}-\dfrac{5}{4}i

Compact solution

Let $z=-\dfrac{2}{3}+\dfrac{5}{4}i$ . The additive inverse changes both signs, so that its sum with the original number is $0$ .

\begin{aligned}-z &= -(-\dfrac{2}{3}+\dfrac{5}{4}i) \\ &= \dfrac{2}{3}-\dfrac{5}{4}i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$-(a-b)=-a+b$}}\end{aligned}

Question 3

Find the multiplicative inverse of each complex number:

(i)

4+5i

Solution

Compact solution

The multiplicative inverse is the number you multiply $z$ by to get $1$ . It is written $z^{-1}$ and equals $\frac{1}{z}$ .
$z^{-1} = \frac{1}{4+5i}$
The reciprocal is correct; next write it in rectangular form. Multiply the top and the bottom by the conjugate of the bottom. The conjugate of $4+5i$ is $4-5i$ .
$z^{-1} = \frac{1}{4+5i}\times\frac{4-5i}{4-5i}$
Take the denominator first. It is of the form $(a+b)(a-b)$ , which is always $a^{2}-b^{2}$ .
$\left(4+5i\right)\left(4-5i\right) = 4^{2}-\left(5i\right)^{2}$
Square each part. Because $i^{2}=-1$ , the second square is negative, so subtracting it adds.
$\left(4+5i\right)\left(4-5i\right) = 4^{2}+25 = 41$
The numerator is just the conjugate, because the top was $1$ .
$z^{-1} = \frac{4-5i}{41}$
Split it into a real part and an imaginary part.
$z^{-1} = \frac{4}{41}-\frac{5}{41}i$

Answer

\frac{4}{41}-\frac{5}{41}i

Compact solution

Let $z=4+5i$ . For $z=4+5i$ , find $z^{-1}=1/z$ . Write the reciprocal and use the conjugate $4-5i$ .

\begin{aligned}z^{-1} &= \frac{1}{4+5i} \\ &= \frac{1}{4+5i}\cdot\frac{4-5i}{4-5i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{4-5i}{4-5i}=1$}} \\ &= \frac{4-5i}{(4+5i)(4-5i)}\end{aligned}

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(4+5i)(4-5i) &= 4^2-(5i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 16-25i^2 \\ &= 16-25(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 16+25 \\ &= 41\end{aligned}

Substitute the numerator and denominator, then simplify to rectangular form.

\begin{aligned}z^{-1} &= \frac{4-5i}{41} \\ &= \frac{4}{41}-\frac{5}{41}i\end{aligned}

(ii)

6+2i

Solution

Compact solution

The multiplicative inverse is the number you multiply $z$ by to get $1$ . It is written $z^{-1}$ and equals $\frac{1}{z}$ .
$z^{-1} = \frac{1}{6+2i}$
The reciprocal is correct; next write it in rectangular form. Multiply the top and the bottom by the conjugate of the bottom. The conjugate of $6+2i$ is $6-2i$ .
$z^{-1} = \frac{1}{6+2i}\times\frac{6-2i}{6-2i}$
Take the denominator first. It is of the form $(a+b)(a-b)$ , which is always $a^{2}-b^{2}$ .
$\left(6+2i\right)\left(6-2i\right) = 6^{2}-\left(2i\right)^{2}$
Square each part. Because $i^{2}=-1$ , the second square is negative, so subtracting it adds.
$\left(6+2i\right)\left(6-2i\right) = 6^{2}+4 = 40$
The numerator is just the conjugate, because the top was $1$ .
$z^{-1} = \frac{6-2i}{40}$
Split it into a real part and an imaginary part.
$z^{-1} = \frac{6}{40}-\frac{2}{40}i = \frac{3}{20}-\frac{1}{20}i$

Answer

\frac{6}{40}-\frac{2}{40}i = \frac{3}{20}-\frac{1}{20}i

Compact solution

Let $z=6+2i$ . For $z=6+2i$ , find $z^{-1}=1/z$ . Write the reciprocal and use the conjugate $6-2i$ .

\begin{aligned}z^{-1} &= \frac{1}{6+2i} \\ &= \frac{1}{6+2i}\cdot\frac{6-2i}{6-2i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{6-2i}{6-2i}=1$}} \\ &= \frac{6-2i}{(6+2i)(6-2i)}\end{aligned}

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(6+2i)(6-2i) &= 6^2-(2i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 36-4i^2 \\ &= 36-4(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 36+4 \\ &= 40\end{aligned}

Substitute the numerator and denominator, then simplify to rectangular form.

\begin{aligned}z^{-1} &= \frac{6-2i}{40} \\ &= \frac{2(3-i)}{2(20)} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{cancel the common factor }2$}} \\ &= \frac{3-i}{20} \\ &= \frac{3}{20}-\frac{1}{20}i\end{aligned}

Common mistake

Watch outBoth fractions still cancel by

2

. Always reduce the fractions before you stop.

(iii)

7-3i

Solution

Compact solution

The multiplicative inverse is the number you multiply $z$ by to get $1$ . It is written $z^{-1}$ and equals $\frac{1}{z}$ .
$z^{-1} = \frac{1}{7-3i}$
The reciprocal is correct; next write it in rectangular form. Multiply the top and the bottom by the conjugate of the bottom. The conjugate of $7-3i$ is $7+3i$ .
$z^{-1} = \frac{1}{7-3i}\times\frac{7+3i}{7+3i}$
Take the denominator first. It is of the form $(a+b)(a-b)$ , which is always $a^{2}-b^{2}$ .
$\left(7-3i\right)\left(7+3i\right) = 7^{2}-\left(-3i\right)^{2}$
Square each part. Because $i^{2}=-1$ , the second square is negative, so subtracting it adds.
$\left(7-3i\right)\left(7+3i\right) = 7^{2}+9 = 58$
The numerator is just the conjugate, because the top was $1$ .
$z^{-1} = \frac{7+3i}{58}$
Split it into a real part and an imaginary part.
$z^{-1} = \frac{7}{58}+\frac{3}{58}i$

Answer

\frac{7}{58}+\frac{3}{58}i

Compact solution

Let $z=7-3i$ . For $z=7-3i$ , find $z^{-1}=1/z$ . Write the reciprocal and use the conjugate $7+3i$ .

\begin{aligned}z^{-1} &= \frac{1}{7-3i} \\ &= \frac{1}{7-3i}\cdot\frac{7+3i}{7+3i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{7+3i}{7+3i}=1$}} \\ &= \frac{7+3i}{(7-3i)(7+3i)}\end{aligned}

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(7-3i)(7+3i) &= 7^2-(3i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 49-9i^2 \\ &= 49-9(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 49+9 \\ &= 58\end{aligned}

Substitute the numerator and denominator, then simplify to rectangular form.

\begin{aligned}z^{-1} &= \frac{7+3i}{58} \\ &= \frac{7}{58}+\frac{3}{58}i\end{aligned}

(iv)

\sqrt{5}-4i

Solution

Compact solution

The multiplicative inverse is $\frac{1}{z}$ . The surd $\sqrt{5}$ is just an ordinary real number here, so the method does not change.
$z^{-1} = \frac{1}{\sqrt{5}-4i}$
Multiply the top and the bottom by the conjugate of the bottom, which is $\sqrt{5}+4i$ .
$z^{-1} = \frac{1}{\sqrt{5}-4i}\times\frac{\sqrt{5}+4i}{\sqrt{5}+4i}$
Take the denominator, using $(a-b)(a+b)=a^{2}-b^{2}$ with $a=\sqrt{5}$ and $b=4i$ .
$(\sqrt{5}-4i)(\sqrt{5}+4i) = (\sqrt{5})^{2}-(4i)^{2}$
Squaring a square root gives back the number inside, so $(\sqrt{5})^{2}=5$ .
$(\sqrt{5})^{2} = 5$
For the second square, square both the $4$ and the $i$ , and use $i^{2}=-1$ .
$(4i)^{2} = 16i^{2} = -16$
Subtract. Taking away $-16$ is the same as adding $16$ .
$(\sqrt{5}-4i)(\sqrt{5}+4i) = 5+16 = 21$
The numerator is the conjugate, because the top was $1$ .
$z^{-1} = \frac{\sqrt{5}+4i}{21}$
Split into a real part and an imaginary part. The surd stays as it is, because $5$ is not a perfect square.
$z^{-1} = \frac{\sqrt{5}}{21}+\frac{4}{21}i$

Answer

\dfrac{\sqrt{5}}{21}+\dfrac{4}{21}i

Compact solution

Let $z=\sqrt{5}-4i$ . Use the conjugate $\sqrt5+4i$ to make the denominator real.

\begin{aligned}z^{-1} &= \frac1{\sqrt5-4i} \\ &= \frac1{\sqrt5-4i}\cdot\frac{\sqrt5+4i}{\sqrt5+4i} \\ &= \frac{\sqrt5+4i}{(\sqrt5-4i)(\sqrt5+4i)}\end{aligned}

Calculate the denominator first.

\begin{aligned}(\sqrt5-4i)(\sqrt5+4i) &= (\sqrt5)^2-(4i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a-b)(a+b)=a^2-b^2$}} \\ &= 5-16i^2 \\ &= 5-16(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 5+16 \\ &= 21\end{aligned}

Substitute the denominator and separate the two coefficients.

\begin{aligned}z^{-1} &= \frac{\sqrt5+4i}{21} \\ &= \frac{\sqrt5}{21}+\frac4{21}i\end{aligned}

Question 4

If $z_{1}=2+5i$ , $z_{2}=1-3i$ and $z_{3}=2+i$ , then verify that:

(i)

z_{1}+z_{2}=z_{2}+z_{1}

Solution

Compact solution

To verify means to work out both sides separately and show they agree. Never assume they are equal. Start with the left side.
$z_{1}+z_{2} = (2+5i)+(1-3i)$
Add the real parts and the imaginary parts separately.
$z_{1}+z_{2} = (2+1)+(5-3)i = 3+2i$
Now the right side, with the two numbers the other way round.
$z_{2}+z_{1} = (1-3i)+(2+5i)$
Again add the parts separately.
$z_{2}+z_{1} = (1+2)+(-3+5)i = 3+2i$
Both sides come to the same number, which is what was to be shown. This is the commutative property of addition.
$z_{1}+z_{2} = z_{2}+z_{1} = 3+2i$

AnswerBoth sides give

3+2i

, so the property holds.

Compact solution

Verify by working each side out on its own. The left side first.

\begin{aligned}z_{1}+z_{2} &= (2+5i)+(1-3i) \\ &= (2+1)+(5-3)i \\ &= 3+2i\end{aligned}

Now the right side, with the two numbers the other way round.

\begin{aligned}z_{2}+z_{1} &= (1-3i)+(2+5i) \\ &= (1+2)+(-3+5)i \\ &= 3+2i\end{aligned}

Both sides come to the same number, which is the commutative property of addition.

\begin{aligned}z_{1}+z_{2} &= z_{2}+z_{1} = 3+2i\end{aligned}

(ii)

z_{1}z_{2}=z_{2}z_{1}

Solution

Compact solution

Work out the left side first, multiplying the brackets out term by term.
$z_{1}z_{2} = (2+5i)(1-3i) = 2-6i+5i-15i^{2}$
Replace $i^{2}$ by $-1$ . Note $-15(-1)=+15$ , because minus times minus gives plus.
$-15i^{2} = 15$
Collect the real numbers and the $i$ terms.
$z_{1}z_{2} = (2+15)+(-6+5)i = 17-i$
Now the right side, with the brackets the other way round.
$z_{2}z_{1} = (1-3i)(2+5i) = 2+5i-6i-15i^{2}$
Again replace $i^{2}$ by $-1$ .
$-15i^{2} = 15$
Collect the real numbers and the $i$ terms separately.
$z_{2}z_{1} = (2+15)+(5-6)i = 17-i$
Both sides agree. This is the commutative property of multiplication.
$z_{1}z_{2} = z_{2}z_{1} = 17-i$

AnswerBoth sides give

17-i

, so the property holds.

Compact solution

The left side first, multiplying out term by term.

\begin{aligned}z_{1}z_{2} &= (2+5i)(1-3i) \\ &= 2-6i+5i-15i^{2} \\ &= 2-i-15(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^{2}=-1$}} \\ &= 17-i\end{aligned}

Now the right side, with the brackets the other way round.

\begin{aligned}z_{2}z_{1} &= (1-3i)(2+5i) \\ &= 2+5i-6i-15i^{2} \\ &= 2-i-15(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^{2}=-1$}} \\ &= 17-i\end{aligned}

Both sides agree, which is the commutative property of multiplication.

\begin{aligned}z_{1}z_{2} &= z_{2}z_{1} = 17-i\end{aligned}

(iii)

(z_{1}+z_{2})+z_{3}=z_{1}+(z_{2}+z_{3})

Solution

Compact solution

On the left, the brackets say to add $z_{1}$ and $z_{2}$ first. Do the inside of the bracket before anything else.
$z_{1}+z_{2} = (2+1)+(5-3)i = 3+2i$
Now add $z_{3}$ to that result.
$(z_{1}+z_{2})+z_{3} = (3+2i)+(2+i) = (3+2)+(2+1)i = 5+3i$
On the right, the brackets say to add $z_{2}$ and $z_{3}$ first instead.
$z_{2}+z_{3} = (1+2)+(-3+1)i = 3-2i$
Now add $z_{1}$ to that result.
$z_{1}+(z_{2}+z_{3}) = (2+5i)+(3-2i) = (2+3)+(5-2)i = 5+3i$
Both sides agree, so the grouping made no difference. This is the associative property of addition.
$(z_{1}+z_{2})+z_{3} = z_{1}+(z_{2}+z_{3}) = 5+3i$

AnswerBoth sides give

5+3i

, so the property holds.

Compact solution

On the left the bracket says to add $z_{1}$ and $z_{2}$ first.

\begin{aligned}z_{1}+z_{2} &= (2+1)+(5-3)i \\ &= 3+2i\end{aligned}

Then add $z_{3}$ to that result.

\begin{aligned}(z_{1}+z_{2})+z_{3} &= (3+2i)+(2+i) \\ &= (3+2)+(2+1)i \\ &= 5+3i\end{aligned}

On the right the bracket says to add $z_{2}$ and $z_{3}$ first instead.

\begin{aligned}z_{2}+z_{3} &= (1+2)+(-3+1)i \\ &= 3-2i\end{aligned}

Then add $z_{1}$ .

\begin{aligned}z_{1}+(z_{2}+z_{3}) &= (2+5i)+(3-2i) \\ &= (2+3)+(5-2)i \\ &= 5+3i\end{aligned}

The grouping made no difference, which is the associative property of addition.

\begin{aligned}(z_{1}+z_{2})+z_{3} &= z_{1}+(z_{2}+z_{3}) = 5+3i\end{aligned}

(iv)

(z_{1}z_{2})z_{3}=z_{1}(z_{2}z_{3})

Solution

Compact solution

On the left, multiply $z_{1}$ and $z_{2}$ first. This was found in part (ii).
$z_{1}z_{2} = 17-i$
Now multiply that by $z_{3}$ , term by term.
$(z_{1}z_{2})z_{3} = (17-i)(2+i) = 34+17i-2i-i^{2}$
Replace $i^{2}$ by $-1$ . Note $-i^{2}=-(-1)=+1$ .
$-i^{2} = 1$
Collect the real numbers and the $i$ terms separately.
$(z_{1}z_{2})z_{3} = (34+1)+(17-2)i = 35+15i$
On the right, multiply $z_{2}$ and $z_{3}$ first instead.
$z_{2}z_{3} = (1-3i)(2+i) = 2+i-6i-3i^{2}$
Replace $i^{2}$ by $-1$ , so $-3i^{2}=+3$ , then collect.
$z_{2}z_{3} = (2+3)+(1-6)i = 5-5i$
Now multiply $z_{1}$ by that result.
$z_{1}(z_{2}z_{3}) = (2+5i)(5-5i) = 10-10i+25i-25i^{2}$
Replace $i^{2}$ by $-1$ , so $-25i^{2}=+25$ , then collect.
$z_{1}(z_{2}z_{3}) = (10+25)+(-10+25)i = 35+15i$
Both sides agree. This is the associative property of multiplication.
$(z_{1}z_{2})z_{3} = z_{1}(z_{2}z_{3}) = 35+15i$

AnswerBoth sides give

35+15i

, so the property holds.

Compact solution

On the left, multiply $z_{1}$ and $z_{2}$ first. That product was found in part (ii).

\begin{aligned}z_{1}z_{2} &= 17-i\end{aligned}

Now multiply that by $z_{3}$ .

\begin{aligned}(z_{1}z_{2})z_{3} &= (17-i)(2+i) \\ &= 34+17i-2i-i^{2} \\ &= 34+15i-(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^{2}=-1$}} \\ &= 35+15i\end{aligned}

On the right, multiply $z_{2}$ and $z_{3}$ first instead.

\begin{aligned}z_{2}z_{3} &= (1-3i)(2+i) \\ &= 2+i-6i-3i^{2} \\ &= 2-5i-3(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^{2}=-1$}} \\ &= 5-5i\end{aligned}

Then multiply $z_{1}$ by that.

\begin{aligned}z_{1}(z_{2}z_{3}) &= (2+5i)(5-5i) \\ &= 10-10i+25i-25i^{2} \\ &= 10+15i-25(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^{2}=-1$}} \\ &= 35+15i\end{aligned}

Both sides agree, which is the associative property of multiplication.

\begin{aligned}(z_{1}z_{2})z_{3} &= z_{1}(z_{2}z_{3}) = 35+15i\end{aligned}

(v)

z_{1}+(-z_{1})=(-z_{1})+z_{1}=0

Solution

Compact solution

First write down $-z_{1}$ . Put a minus in front of the whole number, which changes the sign of both parts.
$-z_{1} = -(2+5i) = -2-5i$
Now add them in the first order, part by part.
$z_{1}+(-z_{1}) = (2+5i)+(-2-5i) = (2-2)+(5-5)i$
Each bracket comes to $0$ , because a number plus its own negative is always $0$ .
$z_{1}+(-z_{1}) = 0+0i = 0$
Now the other order.
$(-z_{1})+z_{1} = (-2-5i)+(2+5i) = (-2+2)+(-5+5)i = 0$
Both orders give $0$ , which is what the additive inverse is defined to do.
$z_{1}+(-z_{1}) = (-z_{1})+z_{1} = 0$

AnswerBoth orders give

0

, so

-z_{1}

is the additive inverse of

z_{1}

Compact solution

First write $-z_{1}$ , which changes the sign of both parts.

\begin{aligned}-z_{1} &= -(2+5i) \\ &= -2-5i\end{aligned}

Add them in the first order, part by part.

\begin{aligned}z_{1}+(-z_{1}) &= (2+5i)+(-2-5i) \\ &= (2-2)+(5-5)i \\ &= 0\end{aligned}

Now the other order.

\begin{aligned}(-z_{1})+z_{1} &= (-2-5i)+(2+5i) \\ &= (-2+2)+(-5+5)i \\ &= 0\end{aligned}

Both orders give $0$ , which is exactly what an additive inverse is for.

\begin{aligned}z_{1}+(-z_{1}) &= (-z_{1})+z_{1} = 0\end{aligned}

Question 5

If $\frac{(1+i)^{2}}{2-i}=x+iy$ , find the values of $x$ and $y$ . Assume $x$ and $y$ are real.

Solution

Compact solution

Deal with the numerator first, because it has a power on a bracket. Use the identity $(a+b)^{2}=a^{2}+2ab+b^{2}$ with $a=1$ and $b=i$ .
$(1+i)^{2} = 1^{2}+2(1)(i)+i^{2}$
Work out each of the three terms.
$1^{2} = 1, \quad 2(1)(i) = 2i, \quad i^{2} = -1$
Put them together and collect the real numbers.
$(1+i)^{2} = 1+2i-1 = 2i$
So the whole expression becomes this.
$\frac{(1+i)^{2}}{2-i} = \frac{2i}{2-i}$
There is an $i$ in the denominator. Multiply the top and the bottom by the conjugate of the bottom, which is $2+i$ .
$\frac{2i}{2-i} = \frac{2i}{2-i}\times\frac{2+i}{2+i}$
Take the denominator, using $(a-b)(a+b)=a^{2}-b^{2}$ .
$(2-i)(2+i) = 2^{2}-i^{2} = 4-(-1) = 5$
Now the numerator.
$2i(2+i) = 4i+2i^{2}$
Replace $i^{2}$ by $-1$ , so $2i^{2}=-2$ .
$2i(2+i) = -2+4i$
Put the numerator over the denominator.
$\frac{(1+i)^{2}}{2-i} = \frac{-2+4i}{5}$
Split it into a real part and an imaginary part, which is the form $x+iy$ .
$\frac{(1+i)^{2}}{2-i} = -\frac{2}{5}+\frac{4}{5}i$
Two complex numbers are equal only when their real parts match and their imaginary parts match. Compare with $x+iy$ .
$x = -\frac{2}{5}, \qquad y = \frac{4}{5}$

Answer

x=-\dfrac{2}{5}

and

y=\dfrac{4}{5}

Compact solution

Deal with the numerator first, because it carries a power on a bracket.

\begin{aligned}(1+i)^{2} &= 1^{2}+2(1)(i)+i^{2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)^{2}=a^{2}+2ab+b^{2}$}} \\ &= 1+2i+(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^{2}=-1$}} \\ &= 2i\end{aligned}

So the whole expression is a single fraction. Clear the $i$ with the conjugate $2+i$ .

\begin{aligned}\frac{(1+i)^{2}}{2-i} &= \frac{2i}{2-i} \\ &= \frac{2i}{2-i}\times\frac{2+i}{2+i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{multiply by }\frac{2+i}{2+i}=1$}} \\ &= \frac{4i+2i^{2}}{2^{2}-i^{2}} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a-b)(a+b)=a^{2}-b^{2}$}} \\ &= \frac{4i-2}{4+1} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^{2}=-1$}} \\ &= \frac{-2+4i}{5} \\ &= -\frac{2}{5}+\frac{4}{5}i\end{aligned}

Comparing that with $x+iy$ , part for part:

\begin{aligned}x &= -\frac{2}{5}, \qquad y = \frac{4}{5}\end{aligned}

Common mistake

Watch outDo not write

(1+i)^{2}=1^{2}+i^{2}

. Squaring a bracket is never done term by term. The middle term

2ab

is what makes this question work out to

2i

Question 6

If $(2x+iy)(1-i)=4+2i$ , find the values of $x$ and $y$ . Assume $x$ and $y$ are real.

Solution

Compact solution

The left side is not yet split into a real part and an imaginary part, so multiply the brackets out first, term by term.
$(2x+iy)(1-i) = 2x(1)+2x(-i)+iy(1)+iy(-i)$
Work out the first three products.
$2x(1) = 2x, \quad 2x(-i) = -2xi, \quad iy(1) = iy$
Now the fourth. Plus times minus gives minus, so this term is $-i^{2}y$ .
$iy(-i) = -i^{2}y$
Replace $i^{2}$ by $-1$ . A minus in front of a minus gives a plus, so this term becomes $+y$ , with no $i$ on it.
$-i^{2}y = -(-1)y = y$
Put the four terms together.
$(2x+iy)(1-i) = 2x-2xi+iy+y$
Gather the terms with no $i$ , and take $i$ out of the other two as a common factor.
$(2x+y)+(y-2x)i = 4+2i$
Two complex numbers are equal only when their real parts match and their imaginary parts match. Compare the real parts and label the equation.
$2x+y = 4 \qquad \cdots(1)$
Now compare the imaginary parts.
$y-2x = 2 \qquad \cdots(2)$
Add equation $(1)$ to equation $(2)$ , left side to left side and right side to right side. We choose to add because the $x$ terms are opposites and will disappear.
$(2x+y)+(y-2x) = 4+2$
On the left, $2x$ and $-2x$ cancel, and $y+y=2y$ . On the right, $4+2=6$ .
$2y = 6$
Divide both sides by $2$ .
$y = 3$
Put $y=3$ back into equation $(1)$ .
$2x+3 = 4$
Take $3$ from both sides.
$2x = 1$
Divide both sides by $2$ .
$x = \frac{1}{2}$

Answer

x=\dfrac{1}{2}

and

y=3

Compact solution

Multiply out and regroup so the left side is in the form $(\text{real})+(\text{real})i$ .

\begin{aligned}(2x+iy)(1-i) &= 2x(1)+2x(-i)+iy(1)+iy(-i) \\ &= 2x-2xi+iy-i^{2}y \\ &= 2x-2xi+iy+y && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^{2}=-1$}} \\ &= (2x+y)+(y-2x)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{taking $i$ out as a common factor}$}}\end{aligned}

Comparing the real and imaginary parts gives two equations.

\begin{aligned}2x+y &= 4\qquad\cdots(1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{real parts}$}} \\ y-2x &= 2\qquad\cdots(2) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{imaginary parts}$}}\end{aligned}

The $x$ terms are opposites, so adding the two equations removes $x$ at once.

\begin{aligned}(2x+y)+(y-2x) &= 4+2 \\ 2y &= 6 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{the $x$ terms cancel}$}} \\ \frac{2y}{2} &= \frac62 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{divide both sides by $2$}$}} \\ y &= 3\end{aligned}

Put that value back into equation $(1)$ .

\begin{aligned}2x+3 &= 4 \\ 2x &= 1 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{take $3$ from both sides}$}} \\ \frac{2x}{2} &= \frac12 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{divide both sides by $2$}$}} \\ x &= \frac12\end{aligned}

Check

Check the answer

\begin{aligned}\left(2\left(\tfrac{1}{2}\right)+3i\right)(1-i) &= (1+3i)(1-i) \\ &= 1-i+3i-3i^{2} \\ &= 1+2i+3 \\ &= 4+2i \quad \checkmark\end{aligned}

Substituting both values into the left side gives the right side exactly.

Question 7

Find the values of $a$ and $b$ , if $(a+bi)(1+3i)=-8+11i$ . Assume $a$ and $b$ are real.

Solution

Compact solution

Multiply the brackets out first, term by term.
$(a+bi)(1+3i) = a(1)+a(3i)+bi(1)+bi(3i)$
Work out the first three products.
$a(1) = a, \quad a(3i) = 3ai, \quad bi(1) = bi$
Now the fourth. It carries $i^{2}$ .
$bi(3i) = 3bi^{2}$
Replace $i^{2}$ by $-1$ , so this term becomes $-3b$ , a real number with no $i$ on it.
$3bi^{2} = -3b$
Put the four terms together.
$(a+bi)(1+3i) = a+3ai+bi-3b$
Gather the terms with no $i$ , and take $i$ out of the other two.
$(a-3b)+(3a+b)i = -8+11i$
Two complex numbers are equal only when both parts match. Compare the real parts.
$a-3b = -8 \qquad \cdots(1)$
Now compare the imaginary parts.
$3a+b = 11 \qquad \cdots(2)$
To eliminate $a$ , multiply equation $(1)$ by $3$ , so that both equations have $3a$ .
$3a-9b = -24 \qquad \cdots(3)$
Now subtract equation $(3)$ from equation $(2)$ , so the $3a$ terms cancel.
$(3a+b)-(3a-9b) = 11-(-24)$
On the left, $3a-3a=0$ , and $b-(-9b)=b+9b=10b$ . On the right, subtracting $-24$ is the same as adding $24$ .
$10b = 35$
Divide both sides by $10$ , then cancel the common factor $5$ .
$b = \frac{35}{10} = \frac{7}{2}$
Put $b=\frac{7}{2}$ back into equation $(2)$ .
$3a+\frac{7}{2} = 11$
Take $\frac{7}{2}$ from both sides. Write $11$ as $\frac{22}{2}$ so the fractions have the same bottom number.
$3a = \frac{22}{2}-\frac{7}{2} = \frac{15}{2}$
Divide both sides by $3$ .
$a = \frac{15}{6} = \frac{5}{2}$

Answer

a=\dfrac{5}{2}

and

b=\dfrac{7}{2}

Compact solution

Multiply out and regroup into a real part and an imaginary part.

\begin{aligned}(a+bi)(1+3i) &= a(1)+a(3i)+bi(1)+bi(3i) \\ &= a+3ai+bi+3bi^{2} \\ &= a+3ai+bi-3b && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^{2}=-1$}} \\ &= (a-3b)+(3a+b)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{taking $i$ out as a common factor}$}}\end{aligned}

Comparing the two parts gives a pair of equations.

\begin{aligned}a-3b &= -8\qquad\cdots(1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{real parts}$}} \\ 3a+b &= 11\qquad\cdots(2) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{imaginary parts}$}}\end{aligned}

Multiply equation $(1)$ by $3$ , so that both equations carry $3a$ .

\begin{aligned}3a-9b &= -24 \qquad\cdots(3)\end{aligned}

Subtract $(3)$ from $(2)$ , so the $3a$ terms cancel.

\begin{aligned}(3a+b)-(3a-9b) &= 11-(-24) \\ 10b &= 35 \\ b &= \frac{35}{10} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{divide both sides by }10$}} \\ &= \frac72 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{cancel the common factor }5$}}\end{aligned}

Put that value back into equation $(2)$ .

\begin{aligned}3a+\frac{7}{2} &= 11 \\ 3a &= \frac{22}{2}-\frac{7}{2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{write $11$ as $\frac{22}{2}$}$}} \\ &= \frac{15}{2} \\ a &= \frac{15}{2}\div3 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{divide both sides by }3$}} \\ &= \frac{15}{6} \\ &= \frac52 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{cancel the common factor }3$}}\end{aligned}

Check

Check the answer

\begin{aligned}\left(\tfrac{5}{2}+\tfrac{7}{2}i\right)(1+3i) &= \tfrac{5}{2}+\tfrac{15}{2}i+\tfrac{7}{2}i+\tfrac{21}{2}i^{2} \\ &= \left(\tfrac{5}{2}-\tfrac{21}{2}\right)+\left(\tfrac{15}{2}+\tfrac{7}{2}\right)i \\ &= -8+11i \quad \checkmark\end{aligned}

Substituting both values reproduces the right-hand side exactly.

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Class 10 Mathematics

Explore the book

Complex Numbers

Quadratic Equations and Inequalities

Matrices and Determinants

Functions and Graphs

Algebraic Fractions

Vectors in Plane

Trigonometry

Chords and Arcs of a Circle

Tangent and Angles of a Circle

Practical Geometry of Circles

Information Handling

Probability

Multiplication by a real number

Adding and subtracting complex numbers

Multiplying complex numbers

Dividing complex numbers

Additive inverse

Multiplicative inverse

Properties of complex numbers

Addition and subtraction on the complex plane

Why multiplying is just ordinary algebra

Why the conjugate clears the denominator

The two inverses are different things

Which algebraic rules still apply?

Example 3

Example 4

Example 5

Example 6

Example 7

Example 8

Example 9

Example 10

Skilled practice

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Class 10 Mathematics

Explore the book

Complex Numbers

Quadratic Equations and Inequalities

Matrices and Determinants

Functions and Graphs

Algebraic Fractions

Vectors in Plane

Trigonometry

Chords and Arcs of a Circle

Tangent and Angles of a Circle

Practical Geometry of Circles

Information Handling

Probability

Class 10
Mathematics

Class 10
Mathematics