Class 10 Review Exercise Solutions — Complex Numbers

Multiple choice

The review exercise covers the whole unit. Question 1 is multiple choice, which is how the objective part of the paper is set. The rest is written work.

Question 1

Four possible answers are given for each question. Choose the correct answer.

(i)

i^{2}+i^{4}=

Solution

Take the two powers one at a time. The first is the definition of $i$ itself.
$i^{2} = -1$
For the second, write it as the square of $i^{2}$ , then use the definition.
$i^{4} = \left(i^{2}\right)^{2} = (-1)^{2} = 1$
Now add the two values. Adding $1$ to $-1$ brings you back to $0$ .
$i^{2}+i^{4} = -1+1 = 0$

Answer

0

Compact solution

Use $i^2=-1$ to evaluate the fourth power.

\begin{aligned}i^4 &= (i^2)^2 \\ &= (-1)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 1 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=1\text{ for even }k$}}\end{aligned}

Add the two powers.

\begin{aligned}i^2+i^4 &= -1+1 \\ &= 0 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a+(-a)=0$}}\end{aligned}

(ii)The real part of

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

is:

Solution

Multiply the two brackets first. They are of the form $(a-b)(a+b)$ , which always gives $a^{2}-b^{2}$ . Here $a=2$ and $b=3i$ .
$(2-3i)(2+3i) = 2^{2}-(3i)^{2}$
Work out each square. For the second, square both the $3$ and the $i$ .
$2^{2} = 4 \quad\text{and}\quad (3i)^{2} = 3^{2}\cdot i^{2} = 9\cdot(-1) = -9$
Subtract. Taking away $-9$ is the same as adding $9$ .
$(2-3i)(2+3i) = 4-(-9) = 4+9 = 13$
The result is $13+0i$ , so its real part is $13$ .
$\operatorname{Re}\bigl((2-3i)(2+3i)\bigr) = 13$

Answer

13

Compact solution

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(2-3i)(2+3i) &= 2^2-(3i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 4-9i^2 \\ &= 4-9(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 4+9 \\ &= 13\end{aligned}

The product is $13=13+0i$ , so its real part is $13$ .

\operatorname{Re}\bigl((2-3i)(2+3i)\bigr)=13

(iii)The imaginary part of

(2-i)(2+i)

is:

Solution

Multiply the brackets. They are of the form $(a-b)(a+b)$ , which gives $a^{2}-b^{2}$ , with $a=2$ and $b=i$ .
$(2-i)(2+i) = 2^{2}-i^{2}$
Replace $i^{2}$ by $-1$ . Taking away $-1$ is the same as adding $1$ .
$(2-i)(2+i) = 4-(-1) = 5$
The result is $5$ , which is $5+0i$ . The imaginary part is the number multiplying $i$ , and here it is $0$ .
$\operatorname{Im}\bigl((2-i)(2+i)\bigr) = 0$

Answer

0

Compact solution

Multiply the conjugate pair using the difference of squares.

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

The product is real, so its imaginary coefficient is $0$ .

\operatorname{Im}\bigl((2-i)(2+i)\bigr)=0

(iv)

x+iy

is a purely imaginary number when:

Textbook wording: option (b) is printed as $x=0$ . The option below states the nonzero convention used in these notes explicitly.

Solution

A number is purely imaginary when it has no real part at all, so the real part $x$ must be $0$ .
$x = 0$
But that alone is not enough. If $y$ were also $0$ the number would be $0$ , and $0$ is a real number, not a purely imaginary one. So $y$ must not be $0$ .
$y \neq 0$
Both conditions together give the answer.
$x = 0 \quad \text{and} \quad y \neq 0$

Answer

x=0

and

y\neq 0

Compact solution

A purely imaginary number has real coefficient $0$ and a nonzero imaginary coefficient. Therefore

x=0\quad\text{and}\quad y\neq0

The nonzero condition excludes $z=0$ . Zero is real; the convention here excludes it from purely imaginary numbers.

Common mistake

Watch outThe supplied textbook prints option (b) as

x=0

. Here we state

y\neq0

as well, following the convention in these notes that purely imaginary numbers exclude

0

. Some authors include

0

among imaginary numbers; this is a convention, not a new calculation.

(v)What is the additive inverse of

5-2i

Solution

The additive inverse of a number is what you add to it to get $0$ . So we need the number $w$ with this property.
$(5-2i)+w = 0$
Take $5-2i$ from both sides. This means changing the sign of every term.
$w = -(5-2i)$
A minus sign in front of a bracket changes the sign of each term inside. So $5$ becomes $-5$ , and $-2i$ becomes $+2i$ .
$w = -5+2i$

Answer

-5+2i

Compact solution

Let $z=5-2i$ . The additive inverse changes both signs, so that its sum with the original number is $0$ .

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

(vi)What is the multiplicative inverse of

z=1+i

Solution

The multiplicative inverse of $z$ is the number you multiply it by to get $1$ . It is written $\frac{1}{z}$ .
$\frac{1}{z} = \frac{1}{1+i}$
Next write the reciprocal in rectangular form. Multiply top and bottom by the conjugate of the bottom, which is $1-i$ . This is allowed because $\frac{1-i}{1-i}=1$ .
$\frac{1}{1+i} = \frac{1}{1+i}\cdot\frac{1-i}{1-i}$
The bottom is of the form $(a+b)(a-b)=a^{2}-b^{2}$ , with $a=1$ and $b=i$ .
$(1+i)(1-i) = 1^{2}-i^{2} = 1-(-1) = 2$
So the whole fraction becomes this.
$\frac{1}{1+i} = \frac{1-i}{2}$
Split it into a real part and an imaginary part.
$\frac{1}{1+i} = \frac{1}{2}-\frac{1}{2}i$

Answer

\dfrac{1}{2}-\dfrac{1}{2}i

Compact solution

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

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

Multiply the conjugate pair using the difference of squares.

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

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

\begin{aligned}z^{-1} &= \frac{1-i}{2} \\ &= \frac{1}{2}-\frac{1}{2}i\end{aligned}

(vii)If

z=4-3i

, then

z\bar{z}=

Solution

First write down the conjugate. The conjugate keeps the real part and changes the sign of the imaginary part.
$\bar{z} = 4+3i$
Now multiply. The two brackets are of the form $(a-b)(a+b)$ , which gives $a^{2}-b^{2}$ .
$z\bar{z} = (4-3i)(4+3i) = 4^{2}-(3i)^{2}$
Work out the squares, remembering that $i^{2}=-1$ .
$4^{2} = 16 \quad\text{and}\quad (3i)^{2} = 9i^{2} = -9$
Subtract. Taking away $-9$ is the same as adding $9$ .
$z\bar{z} = 16-(-9) = 25$

Answer

25

Compact solution

Keep the real coefficient and change the sign of the imaginary coefficient.

\begin{aligned}\bar z &= \overline{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}

Thus

z\bar z=25

(viii)The conjugate of

9-4i

is:

Solution

To take the conjugate of a complex number, keep the real part exactly as it is and change the sign of the imaginary part only.
$\overline{x+iy} = x-iy$
Here the real part is $9$ and the imaginary part is $-4$ . So the $9$ stays, and $-4$ becomes $+4$ .
$\overline{9-4i} = 9+4i$

Answer

9+4i

Compact solution

Keep the real coefficient and change the sign of the imaginary coefficient.

\begin{aligned}\overline{9-4i} &= 9+4i\end{aligned}

(ix)If

z=4+4i

, then

z+\bar{z}=

Solution

Write down the conjugate first. Keep the real part, change the sign of the imaginary part.
$\bar{z} = 4-4i$
Now add the two numbers. Add the real parts together and the imaginary parts together.
$z+\bar{z} = (4+4i)+(4-4i)$
The real parts give $4+4=8$ . The imaginary parts are opposites, so $4i-4i=0$ .
$z+\bar{z} = 8+0i = 8$

Answer

8

Compact solution

Keep the real coefficient and change the sign of the imaginary coefficient.

\begin{aligned}\bar z &= \overline{4+4i} \\ &= 4-4i\end{aligned}

Add the real coefficients and the imaginary coefficients separately.

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

(x)If

z=5+4i

, then

|z|=

Solution

The modulus of $z=x+iy$ is found from $|z|=\sqrt{x^{2}+y^{2}}$ . Here $x=5$ and $y=4$ .
$|z| = \sqrt{5^{2}+4^{2}}$
Square each part and add, keeping everything under the one root.
$|z| = \sqrt{25+16} = \sqrt{41}$
Now take the square root. Since $41$ is not a perfect square, the root stays as it is.
$|z| = \sqrt{41}$

Answer

\sqrt{41}

Compact solution

The real and imaginary coefficients are $5$ and $4$ . Use the nonnegative square root.

\begin{aligned}\left|5+4i\right| &= \sqrt{5^2+4^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{25+16} \\ &= \sqrt{41}\end{aligned}

Written questions

Question 2

Answer the following in your own words.

(i)Is

0

a complex number? Explain.

Solution

Compact solution

A complex number is any number that can be written as $x+iy$ , where $x$ and $y$ are real numbers. So the test is simply whether $0$ can be written in that shape.
$z = x+iy, \quad x,y \in \mathbb{R}$
It can. Take $x=0$ and $y=0$ .
$0 = 0+0i$
So $0$ is a complex number. Because its imaginary part is $0$ , it is a real number as well. Every real number is a complex number whose imaginary part happens to be $0$ .
$\operatorname{Re}(0)=0, \qquad \operatorname{Im}(0)=0$

AnswerYes.

0=0+0i

, so it is a complex number, and it is also a real number.

Compact solution

A complex number has the form $x+iy$ with $x,y\in\mathbb R$ . Choose $x=0$ and $y=0$ .

0=0+0i

So $0$ is complex. Its imaginary coefficient is $0$ , so it is also real.

\begin{gathered}\operatorname{Re}(0)=0 \\ \operatorname{Im}(0)=0\end{gathered}

Common mistake

Watch out

0

is not purely imaginary. A purely imaginary number needs

x=0

and

y\neq 0

, and here

y=0

(ii)What is the result of multiplying a complex number by its conjugate?

Solution

Compact solution

Let the number be $z=x+iy$ , where $x$ and $y$ are real. Its conjugate keeps the real part and changes the sign of the imaginary part.
$z = x+iy, \qquad \bar{z} = x-iy$
Multiply them. The two brackets are of the form $(a+b)(a-b)$ , which always gives $a^{2}-b^{2}$ . Here $a=x$ and $b=iy$ .
$z\bar{z} = (x+iy)(x-iy) = x^{2}-(iy)^{2}$
Square the second term. Square both the $i$ and the $y$ , then use $i^{2}=-1$ .
$(iy)^{2} = i^{2}y^{2} = (-1)y^{2} = -y^{2}$
Subtract. Taking away $-y^{2}$ is the same as adding $y^{2}$ .
$z\bar{z} = x^{2}-(-y^{2}) = x^{2}+y^{2}$
Both $x^{2}$ and $y^{2}$ are real, so the answer is always real. It is also never negative, and it equals the square of the modulus.
$z\bar{z} = x^{2}+y^{2} = |z|^{2}$

AnswerA real number, equal to

x^{2}+y^{2}

, which is

|z|^{2}

Compact solution

Let $z=x+iy$ , where $x,y\in\mathbb R$ . Then $\bar z=x-iy$ .

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

The sum of these real squares is nonnegative and is the square of the modulus.

z\bar z=x^2+y^2=|z|^2\geq0

(iii)State the condition for two complex numbers to be equal.

Solution

Compact solution

Let the two numbers be written in the form $x+iy$ , with all four of $x_{1}$ , $y_{1}$ , $x_{2}$ and $y_{2}$ real.
$z_{1} = x_{1}+iy_{1}, \qquad z_{2} = x_{2}+iy_{2}$
They are equal only when both parts match separately. Matching one part is not enough.
$z_{1}=z_{2} \iff x_{1}=x_{2} \ \text{ and } \ y_{1}=y_{2}$
The same statement using the part names.
$\operatorname{Re}(z_{1})=\operatorname{Re}(z_{2}) \ \text{ and } \ \operatorname{Im}(z_{1})=\operatorname{Im}(z_{2})$

AnswerTheir real parts must be equal and their imaginary parts must be equal.

Compact solution

Write $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ , with all four coefficients real. Equality requires both coefficients to match.

z_1=z_2\iff x_1=x_2\ \text{and}\ y_1=y_2

Common mistake

Watch outThe four numbers must be real for this to work. If they were themselves complex, comparing parts would not be valid.

Question 3

Simplify:

(i)

i^{37}

Solution

Compact solution

The exponent $37$ is odd, so take one factor of $i$ out, leaving the even power $36$ .
$i^{37} = i^{36}\cdot i$
Halve the even exponent, because $36=2\times 18$ .
$i^{37} = \left(i^{2}\right)^{18}\cdot i$
Put $-1$ in place of $i^{2}$ . $18$ minus signs cancel in pairs, so the result is $1$ .
$i^{37} = (-1)^{18}\cdot i = i$

Answer

i

Compact solution

Take out one factor of $i$ , then use $i^2=-1$ .

\begin{aligned}i^{37} &= i^{36}\cdot i \\ &= (i^2)^{18}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a^{mn}=(a^m)^n$}} \\ &= (-1)^{18}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 1\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=1\text{ for even }k$}} \\ &= i\end{aligned}

(ii)

i^{13}\times i^{11}

Solution

Compact solution

Both powers have the same base $i$ , so join them into one. When powers of the same base are multiplied, the powers add.
$i^{13}\cdot i^{11} = i^{13+11}$
Do the addition.
$13+11 = 24$
The exponent $24$ is even, so halve it and write the power using $i^{2}$ .
$i^{24} = \left(i^{2}\right)^{12}$
Put $-1$ in place of $i^{2}$ . $12$ minus signs cancel in pairs, so the result is $1$ .
$i^{24} = (-1)^{12} = 1$

Answer

1

Compact solution

The bases are the same, so add the exponents.

\begin{aligned}i^{13}\cdot i^{11} &= i^{13+11} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a^m a^n=a^{m+n}$}} \\ &= i^{24}\end{aligned}

The exponent is even, so write the power in terms of $i^2$ .

\begin{aligned}i^{24} &= (i^2)^{12} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a^{mn}=(a^m)^n$}} \\ &= (-1)^{12} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 1 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=1\text{ for even }k$}}\end{aligned}

Thus the original product is

i^{13}\cdot i^{11}=1

(iii)

(-i)^{-9}

Solution

Compact solution

There is a minus sign inside the bracket and a negative power outside. Deal with the negative power first. It means turn the bracket upside down.
$(-i)^{-9} = \frac{1}{(-i)^{9}}$
Now the bottom. A minus sign in front of something means that thing times $-1$ , so raise each factor to the ninth power on its own.
$(-i)^{9} = (-1)^{9}\cdot i^{9}$
$9$ copies of $-1$ multiplied together. Minus signs cancel in pairs, and $9$ is odd, so one is left over.
$(-1)^{9} = -1$
For the other factor, the exponent $9$ is odd, so take one $i$ out and halve the rest.
$i^{9} = \left(i^{2}\right)^{4}\cdot i = (-1)^{4}\cdot i = i$
Multiply the two factors.
$(-i)^{9} = (-1)\cdot i = -i$
Put that back at the bottom.
$(-i)^{-9} = \frac{1}{-i}$
Next write the reciprocal in rectangular form. Multiply top and bottom by $i$ , then use $i^{2}=-1$ . A minus in front of a minus gives a plus.
$\begin{aligned}(-i)^{-9} &= \frac{1}{-i}\cdot\frac{i}{i} \\ &= \frac{i}{-i^{2}} \\ &= \frac{i}{-(-1)} \\ &= i\end{aligned}$

Answer

i

Compact solution

First rewrite the negative power as a reciprocal.

(-i)^{-9}=\frac{1}{(-i)^{9}}

In the denominator, $-i=(-1)\cdot i$ . Separate the two factors.

(-i)^{9}=(-1)^{9}i^{9}

Take out one factor of $i$ , then use $i^2=-1$ .

\begin{aligned}i^{9} &= i^{8}\cdot i \\ &= (i^2)^{4}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$a^{mn}=(a^m)^n$}} \\ &= (-1)^{4}\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 1\cdot i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=1\text{ for even }k$}} \\ &= i\end{aligned}

Since $9$ is odd, $(-1)^{9}=-1$ . Substitute both reduced factors.

\begin{aligned}(-i)^{9} &= (-1)(i) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(-1)^k=-1\text{ for odd }k$}} \\ &= -i\end{aligned}

Now substitute into the reciprocal and simplify.

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

(iv)

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

Solution

Compact solution

Multiply the brackets term by term. Each term in the first bracket multiplies each term in the second, giving four products.
$(3-4i)(5-6i) = 3\cdot 5 + 3\cdot(-6i) + (-4i)\cdot 5 + (-4i)\cdot(-6i)$
Work out each product. In the last one, minus times minus gives plus, and $4\cdot 6=24$ .
$15 - 18i - 20i + 24i^{2}$
Replace $i^{2}$ by $-1$ in the last term.
$24i^{2} = 24(-1) = -24$
Now collect the real numbers together and the $i$ terms together.
$(15-24) + (-18-20)i$
Taking $24$ from $15$ goes $9$ 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}

(v)

(3+4i)\div(5-7i)

Solution

Compact solution

Division is written as a fraction.
$(3+4i)\div(5-7i) = \frac{3+4i}{5-7i}$
There is $i$ at the bottom, so this is not a finished answer. Multiply top and bottom by the conjugate of the bottom, which is $5+7i$ . This is allowed because that fraction equals $1$ .
$\frac{3+4i}{5-7i} = \frac{3+4i}{5-7i}\cdot\frac{5+7i}{5+7i}$
Take the bottom first. It is of the form $(a-b)(a+b)=a^{2}-b^{2}$ , and $i^{2}=-1$ turns the subtraction into an addition.
$(5-7i)(5+7i) = 5^{2}-(7i)^{2} = 25-49(-1) = 74$
Now the top, multiplying term by term.
$(3+4i)(5+7i) = 15+21i+20i+28i^{2}$
Replace $i^{2}$ by $-1$ and collect. The real numbers give $15-28=-13$ , and the $i$ terms give $21+20=41$ .
$(3+4i)(5+7i) = -13+41i$
Put the top over the bottom.
$\frac{3+4i}{5-7i} = \frac{-13+41i}{74}$

Answer

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

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}

Other forms

Also correct

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

is equally correct and shows the two parts separately.

Question 4

Find the additive and multiplicative inverse of $z=8+9i$ .

Solution

Compact solution

Two different inverses are wanted, so take them one at a time. The additive inverse is the number you add to $z$ to get $0$ , and it is written $-z$ .
$z+(-z) = 0$
To find it, put a minus in front of the whole number. A minus in front of a bracket changes the sign of every term inside.
$-z = -(8+9i)$
Change the sign of both parts. The real part $8$ becomes $-8$ , and the imaginary part $+9$ becomes $-9$ .
$-z = -8-9i$
Now the multiplicative inverse. It is the number you multiply $z$ by to get $1$ , written $z^{-1}$ , and it equals $\frac{1}{z}$ .
$z^{-1} = \frac{1}{8+9i}$
The reciprocal is correct; next write it in rectangular form. Multiply the top and the bottom by the conjugate of the bottom, which is $8-9i$ .
$z^{-1} = \frac{1}{8+9i}\times\frac{8-9i}{8-9i}$
Work out the denominator. The two brackets are of the form $(a+b)(a-b)$ , which is always $a^{2}-b^{2}$ .
$(8+9i)(8-9i) = 8^{2}-(9i)^{2}$
Square each part. Since $(9i)^{2}=81i^{2}=-81$ , taking it away adds $81$ .
$(8+9i)(8-9i) = 64+81 = 145$
The numerator is just $8-9i$ , because the top was $1$ .
$z^{-1} = \frac{8-9i}{145}$
Split it into a real part and an imaginary part.
$z^{-1} = \frac{8}{145}-\frac{9}{145}i$
State both inverses together.
$-z = -8-9i, \qquad z^{-1} = \frac{8}{145}-\frac{9}{145}i$

Answer

-z=-8-9i

and

z^{-1}=\dfrac{8}{145}-\dfrac{9}{145}i

Compact solution

The additive inverse changes both signs, so that its sum with the original number is $0$ .

\begin{aligned}-z &= -(8+9i) \\ &= -8-9i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$-(a+b)=-a-b$}}\end{aligned}

For the multiplicative inverse, use the reciprocal instead. Write the reciprocal and use the conjugate $8-9i$ .

\begin{aligned}z^{-1} &= \frac{1}{8+9i} \\ &= \frac{1}{8+9i}\cdot\frac{8-9i}{8-9i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{8-9i}{8-9i}=1$}} \\ &= \frac{8-9i}{(8+9i)(8-9i)}\end{aligned}

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(8+9i)(8-9i) &= 8^2-(9i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 64-81i^2 \\ &= 64-81(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 64+81 \\ &= 145\end{aligned}

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

\begin{aligned}z^{-1} &= \frac{8-9i}{145} \\ &= \frac{8}{145}-\frac{9}{145}i\end{aligned}

Common mistake

Watch outThese are two different things. The additive inverse needs no division at all, and only the multiplicative inverse needs the conjugate.

Question 5

If $z_{1}=3+4i$ and $z_{2}=2+3i$ , then verify that:

(i)

\overline{z_{1}+z_{2}}=\bar{z_{1}}+\bar{z_{2}}

Solution

Compact solution

To verify, work out each side on its own and show they agree. Never assume they are equal. The left side says add first, then take the conjugate, so add the parts separately.
$z_{1}+z_{2} = (3+2)+(4+3)i = 5+7i$
Now take the conjugate of that sum, keeping the real part and changing the sign of the imaginary part.
$\overline{z_{1}+z_{2}} = 5-7i \qquad \cdots(1)$
The right side says take the conjugates first. Write each one.
$\bar{z_{1}} = 3-4i, \qquad \bar{z_{2}} = 2-3i$
Add them, part by part. The imaginary parts give $-4-3=-7$ , because going down $3$ more from $-4$ reaches $-7$ .
$\bar{z_{1}}+\bar{z_{2}} = (3+2)+(-4-3)i = 5-7i \qquad \cdots(2)$
From $(1)$ and $(2)$ both sides give the same number.
$\overline{z_{1}+z_{2}} = \bar{z_{1}}+\bar{z_{2}} = 5-7i$

AnswerBoth sides give

5-7i

, so the property holds.

Compact solution

Calculate the left side first: perform the operation before conjugating. Add the real coefficients and the imaginary coefficients separately.

\begin{aligned}z_1+z_2 &= (3+4i)+(2+3i) \\ &= 3+4i+2+3i \\ &= (3+2)+(4+3)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 5+7i\end{aligned}

Take the conjugate of this result.

\begin{aligned}\overline{z_1+z_2} &= \overline{5+7i} \\ &= 5-7i\end{aligned}

For the right side, conjugate the two inputs first.

\begin{gathered}\bar z_1=3-4i \\ \bar z_2=2-3i\end{gathered}

Add the real coefficients and the imaginary coefficients separately.

\begin{aligned}\bar z_1+\bar z_2 &= (3-4i)+(2-3i) \\ &= 3-4i+2-3i \\ &= (3+2)+(-4-3)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 5-7i\end{aligned}

Both sides give $5-7i$ , so the required equality holds for these values.

\overline{z_1+z_2}=\bar z_1+\bar z_2=5-7i

(ii)

\overline{z_{1}z_{2}}=\bar{z_{1}}\,\bar{z_{2}}

Solution

Compact solution

The left side says multiply first, then take the conjugate. Multiply the brackets out term by term.
$z_{1}z_{2} = (3+4i)(2+3i) = 6+9i+8i+12i^{2}$
Replace $i^{2}$ by $-1$ , so $12i^{2}=-12$ .
$z_{1}z_{2} = 6+9i+8i-12$
Collect. The real numbers give $6-12=-6$ , and the $i$ terms give $9i+8i=17i$ .
$z_{1}z_{2} = -6+17i$
Now take the conjugate of that product.
$\overline{z_{1}z_{2}} = -6-17i \qquad \cdots(1)$
The right side says take the conjugates first. Write each one.
$\bar{z_{1}} = 3-4i, \qquad \bar{z_{2}} = 2-3i$
Multiply them out term by term.
$\bar{z_{1}}\,\bar{z_{2}} = (3-4i)(2-3i) = 6-9i-8i+12i^{2}$
Replace $i^{2}$ by $-1$ and collect. The real numbers give $6-12=-6$ , and the $i$ terms give $-9i-8i=-17i$ .
$\bar{z_{1}}\,\bar{z_{2}} = -6-17i \qquad \cdots(2)$
From $(1)$ and $(2)$ both sides agree.
$\overline{z_{1}z_{2}} = \bar{z_{1}}\,\bar{z_{2}} = -6-17i$

AnswerBoth sides give

-6-17i

, so the property holds.

Compact solution

Calculate the left side first: perform the operation before conjugating. Multiply each term in the first bracket by each term in the second.

\begin{aligned}z_1z_2 &= (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}{$\text{collect like terms}$}} \\ &= -6+17i\end{aligned}

Take the conjugate of this result.

\begin{aligned}\overline{z_1z_2} &= \overline{-6+17i} \\ &= -6-17i\end{aligned}

For the right side, conjugate the two inputs first.

\begin{gathered}\bar z_1=3-4i \\ \bar z_2=2-3i\end{gathered}

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

\begin{aligned}\bar z_1\bar z_2 &= (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}{$\text{collect like terms}$}} \\ &= -6-17i\end{aligned}

Both sides give $-6-17i$ , so the required equality holds for these values.

\overline{z_1z_2}=\bar z_1\bar z_2=-6-17i

(iii)

\overline{\left(\frac{z_{1}}{z_{2}}\right)}=\frac{\bar{z_{1}}}{\bar{z_{2}}}

Solution

Compact solution

The left side says divide first, then take the conjugate. Begin with the division.
$\frac{z_{1}}{z_{2}} = \frac{3+4i}{2+3i}$
There is an $i$ in the denominator, so multiply the top and the bottom by the conjugate of the bottom, which is $2-3i$ .
$\frac{z_{1}}{z_{2}} = \frac{3+4i}{2+3i}\times\frac{2-3i}{2-3i}$
Work out the numerator, multiplying term by term.
$(3+4i)(2-3i) = 6-9i+8i-12i^{2}$
Replace $i^{2}$ by $-1$ . Note $-12i^{2}=12$ , so the real numbers give $6+12=18$ , and the $i$ terms give $-9i+8i=-i$ .
$(3+4i)(2-3i) = 18-i$
Work out the denominator, of the form $(a+b)(a-b)=a^{2}-b^{2}$ .
$(2+3i)(2-3i) = 2^{2}-(3i)^{2} = 4+9 = 13$
Put the numerator over the denominator and split it into two parts.
$\frac{z_{1}}{z_{2}} = \frac{18-i}{13} = \frac{18}{13}-\frac{1}{13}i$
Now take the conjugate of that, changing the sign of the imaginary part only.
$\overline{\left(\frac{z_{1}}{z_{2}}\right)} = \frac{18}{13}+\frac{1}{13}i \qquad \cdots(1)$
Turn to the right side, which says take the conjugates first, then divide.
$\frac{\bar{z_{1}}}{\bar{z_{2}}} = \frac{3-4i}{2-3i}$
Multiply the top and the bottom by the conjugate of the new bottom, which is $2+3i$ .
$\frac{\bar{z_{1}}}{\bar{z_{2}}} = \frac{3-4i}{2-3i}\times\frac{2+3i}{2+3i}$
Work out the numerator, then replace $i^{2}$ by $-1$ . Here $6+12=18$ and $9i-8i=i$ .
$(3-4i)(2+3i) = 6+9i-8i-12i^{2} = 18+i$
The denominator is the same pair of brackets as before, so it is $13$ again.
$(2-3i)(2+3i) = 13$
So the right side comes to this.
$\frac{\bar{z_{1}}}{\bar{z_{2}}} = \frac{18+i}{13} = \frac{18}{13}+\frac{1}{13}i \qquad \cdots(2)$
From $(1)$ and $(2)$ both sides agree, and this is the common value.
$\overline{\left(\frac{z_{1}}{z_{2}}\right)} = \frac{\bar{z_{1}}}{\bar{z_{2}}} = \frac{18}{13}+\frac{1}{13}i$

AnswerBoth sides give

\dfrac{18}{13}+\dfrac{1}{13}i

, so the property holds.

Compact solution

Calculate the left side first: perform the operation before conjugating. Multiply numerator and denominator by the conjugate $2-3i$ .

\begin{aligned}\frac{z_1}{z_2} &= \frac{3+4i}{2+3i} \\ &= \frac{3+4i}{2+3i}\cdot\frac{2-3i}{2-3i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{2-3i}{2-3i}=1$}} \\ &= \frac{(3+4i)(2-3i)}{(2+3i)(2-3i)}\end{aligned}

Expand the numerator first.

\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}

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(2+3i)(2-3i) &= 2^2-(3i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 4-9i^2 \\ &= 4-9(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 4+9 \\ &= 13\end{aligned}

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

\begin{aligned}\frac{z_1}{z_2} &= \frac{18-i}{13} \\ &= \frac{18}{13}-\frac{1}{13}i\end{aligned}

Take the conjugate of this result.

\begin{aligned}\overline{\frac{z_1}{z_2}} &= \overline{\dfrac{18}{13}-\dfrac{1}{13}i} \\ &= \dfrac{18}{13}+\dfrac{1}{13}i\end{aligned}

For the right side, conjugate the two inputs first.

\begin{gathered}\bar z_1=3-4i \\ \bar z_2=2-3i\end{gathered}

Multiply numerator and denominator by the conjugate $2+3i$ .

\begin{aligned}\frac{\bar z_1}{\bar z_2} &= \frac{3-4i}{2-3i} \\ &= \frac{3-4i}{2-3i}\cdot\frac{2+3i}{2+3i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{2+3i}{2+3i}=1$}} \\ &= \frac{(3-4i)(2+3i)}{(2-3i)(2+3i)}\end{aligned}

Expand the numerator first.

\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}

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(2-3i)(2+3i) &= 2^2-(3i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 4-9i^2 \\ &= 4-9(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 4+9 \\ &= 13\end{aligned}

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

\begin{aligned}\frac{\bar z_1}{\bar z_2} &= \frac{18+i}{13} \\ &= \frac{18}{13}+\frac{1}{13}i\end{aligned}

Both sides give $\dfrac{18}{13}+\dfrac{1}{13}i$ , so the required equality holds for these values.

\overline{\frac{z_1}{z_2}}=\frac{\bar z_1}{\bar z_2}=\dfrac{18}{13}+\dfrac{1}{13}i

(iv)

|z_{1}|=\left|-\bar{z_{1}}\right|

Solution

Compact solution

Start with the plain modulus. The modulus of $z=x+iy$ is $\sqrt{x^{2}+y^{2}}$ , and here $x=3$ and $y=4$ .
$|z_{1}| = \sqrt{3^{2}+4^{2}}$
Square each part, add them, and take the root. Since $5\times 5=25$ , the root is exact.
$|z_{1}| = \sqrt{9+16} = \sqrt{25} = 5 \qquad \cdots(1)$
Now build $-\bar{z_{1}}$ in two moves. First take the conjugate, which changes the sign of the imaginary part only.
$\bar{z_{1}} = 3-4i$
Then put a minus in front of that, which changes the sign of both parts.
$-\bar{z_{1}} = -3+4i$
Take its modulus, with $x=-3$ and $y=4$ .
$\left|-\bar{z_{1}}\right| = \sqrt{(-3)^{2}+4^{2}}$
Square each part. The minus sign disappears, because minus times minus gives plus.
$\left|-\bar{z_{1}}\right| = \sqrt{9+16} = 5 \qquad \cdots(2)$
From $(1)$ and $(2)$ both are the same. Changing signs never changes how far a point is from the origin.
$|z_{1}| = \left|-\bar{z_{1}}\right| = 5$

AnswerBoth sides give

5

, so the property holds.

Compact solution

The real and imaginary coefficients are $3$ and $4$ . Use the nonnegative square root.

\begin{aligned}|z_1| &= \sqrt{3^2+4^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{9+16} \\ &= \sqrt{25} \\ &= 5\end{aligned}

Keep the real coefficient and change the sign of the imaginary coefficient.

\begin{aligned}\bar{z_1} &= \overline{3+4i} \\ &= 3-4i\end{aligned}

Negate the conjugate by changing both signs.

\begin{aligned}-\bar{z_1} &= -(3-4i) \\ &= -3+4i\end{aligned}

The real and imaginary coefficients are $-3$ and $4$ . Use the nonnegative square root.

\begin{aligned}\left|-\bar{z_1}\right| &= \sqrt{(-3)^2+4^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{9+16} \\ &= \sqrt{25} \\ &= 5\end{aligned}

The two distances are equal.

|z_1|=\left|-\bar{z_1}\right|=5

(v)

\bar{\bar{z_{2}}}=z_{2}

Solution

Compact solution

Take the conjugate of $z_{2}$ once. Keep the real part and change the sign of the imaginary part.
$\bar{z_{2}} = 2-3i$
Now take the conjugate again, of the number just found. The real part $2$ stays, and the imaginary part $-3$ changes sign to $+3$ .
$\bar{\bar{z_{2}}} = \overline{2-3i} = 2+3i$
That is the number we started with, so the property holds.
$\bar{\bar{z_{2}}} = 2+3i = z_{2}$

Answer

\bar{\bar{z_{2}}}=2+3i=z_{2}

Compact solution

Keep the real coefficient and change the sign of the imaginary coefficient.

\begin{aligned}\bar{z_2} &= \overline{2+3i} \\ &= 2-3i\end{aligned}

Conjugate once more; the imaginary sign changes back.

\begin{aligned}\overline{\bar{z_2}} &= \overline{2-3i} \\ &= 2+3i \\ &= z_2\end{aligned}

Common mistake

Watch outOn the complex plane this says that reflecting a point in the real axis twice brings it back to where it started.

(vi)

z_{1}\bar{z_{1}}=|z_{1}|^{2}

Solution

Compact solution

Write the conjugate of $z_{1}$ first, changing the sign of the imaginary part.
$\bar{z_{1}} = 3-4i$
Multiply $z_{1}$ by $\bar{z_{1}}$ . The two brackets are of the form $(a+b)(a-b)$ , which is always $a^{2}-b^{2}$ .
$z_{1}\bar{z_{1}} = (3+4i)(3-4i) = 3^{2}-(4i)^{2}$
Open the second square, squaring both the $4$ and the $i$ .
$(4i)^{2} = 16i^{2} = -16$
Subtract. Taking away $-16$ is the same as adding $16$ , so the answer is an ordinary real number.
$z_{1}\bar{z_{1}} = 9+16 = 25 \qquad \cdots(1)$
Now the other side. Find the modulus, using $|z|=\sqrt{x^{2}+y^{2}}$ with $x=3$ and $y=4$ .
$|z_{1}| = \sqrt{9+16} = 5$
Square that modulus, because the right side asks for $|z_{1}|^{2}$ , not $|z_{1}|$ .
$|z_{1}|^{2} = 5^{2} = 25 \qquad \cdots(2)$
From $(1)$ and $(2)$ both sides give the same number.
$z_{1}\bar{z_{1}} = |z_{1}|^{2} = 25$

AnswerBoth sides give

25

, so the property holds.

Compact solution

Keep the real coefficient and change the sign of the imaginary coefficient.

\begin{aligned}\bar{z_1} &= \overline{3+4i} \\ &= 3-4i\end{aligned}

For the left side, multiply the conjugate pair.

\begin{aligned}z_1\bar{z_1} &= (3+4i)(3-4i) \\ &= 3^2-(4i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 9-16i^2 \\ &= 9-16(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 9+16 \\ &= 25\end{aligned}

For the right side, use the modulus of the original number.

\begin{aligned}|z_1| &= \sqrt{3^2+4^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{9+16} \\ &= \sqrt{25} \\ &= 5\end{aligned}

Square the modulus, as the right side requires.

\begin{aligned}|z_1|^2 &= \left(5\right)^2 \\ &= 25\end{aligned}

The two sides agree.

z_1\bar{z_1}=|z_1|^2=25

Common mistake

Watch outThe right side is

|z_{1}|^{2}

, not

|z_{1}|

. There is no square root at the end.

Question 6

If $z_{1}=5+4i$ and $z_{2}=3+2i$ , then find:

(i)

z_{1}z_{2}

Solution

Compact solution

Multiply the brackets out term by term. Each term in the first multiplies each term in the second, which gives four products.
$z_{1}z_{2} = (5+4i)(3+2i) = 15+10i+12i+8i^{2}$
Replace $i^{2}$ by $-1$ , so $8i^{2}=-8$ .
$z_{1}z_{2} = 15+10i+12i-8$
Collect. The real numbers give $15-8=7$ , and the $i$ terms give $10i+12i=22i$ .
$z_{1}z_{2} = 7+22i$

Answer

7+22i

Compact solution

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

\begin{aligned}z_1z_2 &= (5+4i)(3+2i) \\ &= 5(3)+5(2i)+(4i)(3)+(4i)(2i) \\ &= 15+10i+12i+8i^2 \\ &= 15+10i+12i+8(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 15+10i+12i-8 \\ &= (15-8)+(10+12)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 7+22i\end{aligned}

(ii)

\frac{z_{1}}{z_{2}}

Solution

Compact solution

Write the division as a fraction. There is an $i$ in the denominator, so multiply the top and the bottom by the conjugate of the bottom, which is $3-2i$ .
$\frac{z_{1}}{z_{2}} = \frac{5+4i}{3+2i}\times\frac{3-2i}{3-2i}$
Work out the numerator, multiplying term by term.
$(5+4i)(3-2i) = 15-10i+12i-8i^{2}$
Replace $i^{2}$ by $-1$ . Note that $-8i^{2}=8$ , because minus times minus gives plus.
$(5+4i)(3-2i) = 15-10i+12i+8$
Collect. The real numbers give $15+8=23$ , and the $i$ terms give $-10i+12i=2i$ .
$(5+4i)(3-2i) = 23+2i$
Work out the denominator, of the form $(a+b)(a-b)=a^{2}-b^{2}$ .
$(3+2i)(3-2i) = 3^{2}-(2i)^{2} = 9+4 = 13$
Put the numerator over the denominator and split it into two parts.
$\frac{z_{1}}{z_{2}} = \frac{23+2i}{13} = \frac{23}{13}+\frac{2}{13}i$

Answer

\dfrac{23}{13}+\dfrac{2}{13}i

Compact solution

Multiply numerator and denominator by the conjugate $3-2i$ .

\begin{aligned}\frac{z_1}{z_2} &= \frac{5+4i}{3+2i} \\ &= \frac{5+4i}{3+2i}\cdot\frac{3-2i}{3-2i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{3-2i}{3-2i}=1$}} \\ &= \frac{(5+4i)(3-2i)}{(3+2i)(3-2i)}\end{aligned}

Expand the numerator first.

\begin{aligned}(5+4i)(3-2i) &= 5(3)+5(-2i)+(4i)(3)+(4i)(-2i) \\ &= 15-10i+12i-8i^2 \\ &= 15-10i+12i-8(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 15-10i+12i+8 \\ &= (15+8)+(-10+12)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 23+2i\end{aligned}

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(3+2i)(3-2i) &= 3^2-(2i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 9-4i^2 \\ &= 9-4(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 9+4 \\ &= 13\end{aligned}

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

\begin{aligned}\frac{z_1}{z_2} &= \frac{23+2i}{13} \\ &= \frac{23}{13}+\frac{2}{13}i\end{aligned}

(iii)

\bar{z_{1}}\,\bar{z_{2}}

Solution

Compact solution

Write both conjugates first. Keep each real part and change the sign of each imaginary part.
$\bar{z_{1}} = 5-4i, \qquad \bar{z_{2}} = 3-2i$
Multiply them out term by term.
$\bar{z_{1}}\,\bar{z_{2}} = (5-4i)(3-2i) = 15-10i-12i+8i^{2}$
Replace $i^{2}$ by $-1$ , so $8i^{2}=-8$ .
$\bar{z_{1}}\,\bar{z_{2}} = 15-10i-12i-8$
Collect. The real numbers give $15-8=7$ , and the $i$ terms give $-10i-12i=-22i$ .
$\bar{z_{1}}\,\bar{z_{2}} = 7-22i$

Answer

7-22i

Compact solution

Write the conjugates first.

\begin{gathered}\bar z_1=5-4i \\ \bar z_2=3-2i\end{gathered}

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

\begin{aligned}\bar z_1\bar z_2 &= (5-4i)(3-2i) \\ &= 5(3)+5(-2i)+(-4i)(3)+(-4i)(-2i) \\ &= 15-10i-12i+8i^2 \\ &= 15-10i-12i+8(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 15-10i-12i-8 \\ &= (15-8)+(-10-12)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 7-22i\end{aligned}

Alternative method

Alternative methodThis is the conjugate of the answer to part (i), which is exactly what the property

\overline{z_{1}z_{2}}=\bar{z_{1}}\,\bar{z_{2}}

promises.

(iv)

|z_{1}z_{2}|

Solution

Compact solution

The product was worked out in part (i), so start from it. Its real part is $7$ and its imaginary part is $22$ .
$z_{1}z_{2} = 7+22i$
The modulus of $z=x+iy$ is $\sqrt{x^{2}+y^{2}}$ . Put the two parts into that formula.
$|z_{1}z_{2}| = \sqrt{7^{2}+22^{2}}$
Square each part and add, keeping everything under the one root.
$|z_{1}z_{2}| = \sqrt{49+484} = \sqrt{533}$
Take the square root. Since $533$ is not a perfect square, the root must stay as it is.
$|z_{1}z_{2}| = \sqrt{533}$

Answer

\sqrt{533}

Compact solution

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

\begin{aligned}z_1z_2 &= (5+4i)(3+2i) \\ &= 5(3)+5(2i)+(4i)(3)+(4i)(2i) \\ &= 15+10i+12i+8i^2 \\ &= 15+10i+12i+8(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 15+10i+12i-8 \\ &= (15-8)+(10+12)i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{collect like terms}$}} \\ &= 7+22i\end{aligned}

The real and imaginary coefficients are $7$ and $22$ . Use the nonnegative square root.

\begin{aligned}|z_1z_2| &= \sqrt{7^2+22^2} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$|x+iy|=\sqrt{x^2+y^2}$}} \\ &= \sqrt{49+484} \\ &= \sqrt{533}\end{aligned}

Since $533=13\cdot41$ has no square factor greater than $1$ , the exact modulus remains $\sqrt{533}$ .

Other forms

Also correct

\sqrt{41}\times\sqrt{13}

is the same number, because

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

and

41\times 13=533

Question 7

Find the real and imaginary parts of $z=(2+7i)^{-1}$ .

Solution

Compact solution

A power of $-1$ means turn the number over, using $a^{-n}=\frac{1}{a^{n}}$ , so write it as $1$ over the bracket.
$z = (2+7i)^{-1} = \frac{1}{2+7i}$
The parts can only be read from the form $x+iy$ , and there is still an $i$ in the denominator. Multiply the top and the bottom by the conjugate of the bottom, which is $2-7i$ .
$z = \frac{1}{2+7i}\times\frac{2-7i}{2-7i}$
Work out the denominator. The two brackets are of the form $(a+b)(a-b)$ , which is always $a^{2}-b^{2}$ .
$(2+7i)(2-7i) = 2^{2}-(7i)^{2}$
Square each part. Since $(7i)^{2}=49i^{2}=-49$ , taking it away adds $49$ .
$(2+7i)(2-7i) = 4+49 = 53$
The numerator is just $2-7i$ , because the top was $1$ .
$z = \frac{2-7i}{53}$
Split it into a real part and an imaginary part.
$z = \frac{2}{53}-\frac{7}{53}i$
Now the number is in the form $x+iy$ , so read off the two parts.
$\operatorname{Re}(z) = \frac{2}{53}, \qquad \operatorname{Im}(z) = -\frac{7}{53}$

Answer

\mathrm{Re}(z)=\dfrac{2}{53}

and

\mathrm{Im}(z)=-\dfrac{7}{53}

Compact solution

Let $z=(2+7i)^{-1}$ . Write the reciprocal and use the conjugate $2-7i$ .

\begin{aligned}z &= \frac{1}{2+7i} \\ &= \frac{1}{2+7i}\cdot\frac{2-7i}{2-7i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{2-7i}{2-7i}=1$}} \\ &= \frac{2-7i}{(2+7i)(2-7i)}\end{aligned}

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(2+7i)(2-7i) &= 2^2-(7i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 4-49i^2 \\ &= 4-49(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 4+49 \\ &= 53\end{aligned}

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

\begin{aligned}z &= \frac{2-7i}{53} \\ &= \frac{2}{53}-\frac{7}{53}i\end{aligned}

Read the real coefficient and the coefficient of $i$ .

\begin{gathered}\operatorname{Re}(z)=\dfrac{2}{53} \\ \operatorname{Im}(z)=-\dfrac{7}{53}\end{gathered}

Common mistake

Watch outThe imaginary part is

-\frac{7}{53}

, a real number. The

i

is never part of it.

Question 8

Solve the given simultaneous linear equations with complex coefficients for $z$ and $w$ : $\ iz+(2-i)w=4+i$ and $iz+(3+i)w=3+3i$ .

Solution

The given equations are

\begin{aligned}\ iz+(2-i)w&=4+i,\qquad\cdots(1) \\ iz+(3+i)w&=3+3i.\qquad\cdots(2)\end{aligned}

Compact solution

Call the two equations $(1)$ and $(2)$ . Both begin with exactly the same term $iz$ , so taking one from the other removes $z$ straight away. Take equation $(1)$ away from equation $(2)$ .
$\bigl[(3+i)-(2-i)\bigr]w = (3+3i)-(4+i)$
Work out the bracket on the left. A minus in front of a bracket changes both signs, so $3-2=1$ and $i+i=2i$ .
$(3+i)-(2-i) = 1+2i$
Work out the right side the same way. $3-4=-1$ and $3i-i=2i$ .
$(3+3i)-(4+i) = -1+2i$
So the $z$ terms have gone and one unknown is left.
$(1+2i)w = -1+2i$
Divide both sides by $1+2i$ , then clear the $i$ by multiplying the top and the bottom by the conjugate $1-2i$ .
$w = \frac{-1+2i}{1+2i}\times\frac{1-2i}{1-2i}$
Work out the numerator, multiplying term by term.
$(-1+2i)(1-2i) = -1+2i+2i-4i^{2}$
Replace $i^{2}$ by $-1$ . Note $-4i^{2}=4$ , so the real numbers give $-1+4=3$ , and the $i$ terms give $2i+2i=4i$ .
$(-1+2i)(1-2i) = 3+4i$
Work out the denominator, of the form $(a+b)(a-b)=a^{2}-b^{2}$ .
$(1+2i)(1-2i) = 1^{2}-(2i)^{2} = 1+4 = 5$
So $w$ is this fraction. Split it into two parts.
$w = \frac{3+4i}{5} = \frac{3}{5}+\frac{4}{5}i$
Now go back to equation $(1)$ for $z$ . Work out the term $(2-i)w$ first.
$(2-i)w = \frac{(2-i)(3+4i)}{5}$
Multiply the top out term by term.
$(2-i)(3+4i) = 6+8i-3i-4i^{2}$
Replace $i^{2}$ by $-1$ . Note $-4i^{2}=4$ , so the real numbers give $6+4=10$ , and the $i$ terms give $8i-3i=5i$ .
$(2-i)(3+4i) = 10+5i$
Divide each term by $5$ .
$(2-i)w = \frac{10+5i}{5} = 2+i$
Take this away from both sides of equation $(1)$ , so only $iz$ is left.
$iz = (4+i)-(2+i)$
Work out the right side. $4-2=2$ and $i-i=0$ , so the answer is a plain real number.
$iz = 2$
Divide both sides by $i$ , then clear the $i$ from the bottom by multiplying the top and the bottom by $i$ .
$z = \frac{2}{i}\times\frac{i}{i} = \frac{2i}{i^{2}}$
Replace $i^{2}$ by $-1$ . Dividing by $-1$ just changes the sign.
$z = \frac{2i}{-1} = -2i$
State both answers.
$z = -2i, \qquad w = \frac{3}{5}+\frac{4}{5}i$

Answer

z=-2i

and

w=\dfrac{3}{5}+\dfrac{4}{5}i

Compact solution

Label the two original equations.

\begin{aligned}iz+(2-i)w&=4+i\qquad\cdots(1) \\ iz+(3+i)w&=3+3i\qquad\cdots(2)\end{aligned}

Both equations open with the same term $iz$ , so subtracting $(1)$ from $(2)$ removes $z$ at once.

\begin{aligned}\bigl[(3+i)-(2-i)\bigr]w &= (3+3i)-(4+i)\end{aligned}

Clear both brackets. A minus in front changes every sign inside.

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

Divide by $1+2i$ , then clear the $i$ with the conjugate $1-2i$ .

\begin{aligned}w &= \frac{-1+2i}{1+2i} \\ &= \frac{-1+2i}{1+2i}\times\frac{1-2i}{1-2i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{multiply by }\frac{1-2i}{1-2i}=1$}} \\ &= \frac{-1+2i+2i-4i^{2}}{1^{2}-(2i)^{2}} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^{2}-b^{2}$}} \\ &= \frac{-1+4i-4(-1)}{1-4(-1)} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^{2}=-1$}} \\ &= \frac{-1+4i+4}{1+4} \\ &= \frac{3+4i}{5} \\ &= \frac{3}{5}+\frac{4}{5}i\end{aligned}

Now back to equation $(1)$ for $z$ , starting with the term $(2-i)w$ .

\begin{aligned}(2-i)w &= \frac{(2-i)(3+4i)}{5} \\ &= \frac{6+8i-3i-4i^{2}}{5} \\ &= \frac{6+5i-4(-1)}{5} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^{2}=-1$}} \\ &= \frac{10+5i}{5} \\ &= 2+i\end{aligned}

Take that from both sides of equation $(1)$ , leaving $iz$ on its own.

\begin{aligned}iz &= (4+i)-(2+i) \\ &= 2\end{aligned}

Divide by $i$ , then clear the $i$ from the bottom by multiplying top and bottom by $i$ .

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

Thus the solution is

\begin{aligned}z &= -2i, \qquad w = \frac{3}{5}+\frac{4}{5}i\end{aligned}

Check · Alternative method

Check the answer

\begin{aligned}i(-2i)+(2-i)\left(\frac35+\frac45i\right)&=2+(2+i)=4+i\\i(-2i)+(3+i)\left(\frac35+\frac45i\right)&=2+(1+3i)=3+3i\end{aligned}

Both original equations hold after substituting the values of

z

and

w

Alternative methodSubtracting is quicker than substituting here, because the

z

terms are already identical. Look for that before rearranging anything.

Question 9

Solve $(3-4i)(a+bi)=1+0i$ and find the values of $a$ and $b$ . Assume $a$ and $b$ 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 term by term. Here $a$ and $b$ are real numbers.
$(3-4i)(a+bi) = 3a+3bi-4ai-4bi^{2}$
Replace $i^{2}$ by $-1$ . Note that $-4bi^{2}=4b$ , because minus times minus gives plus.
$(3-4i)(a+bi) = 3a+3bi-4ai+4b$
Collect the terms with no $i$ into one bracket and the terms with $i$ into another, taking $i$ out as a common factor.
$(3-4i)(a+bi) = (3a+4b)+(3b-4a)i$
Two complex numbers are equal only when their real parts match and their imaginary parts match. Compare the real parts first, and number the result.
$3a+4b = 1 \qquad \cdots(1)$
Now compare the imaginary parts. The right side is $1+0i$ , so its imaginary part is $0$ .
$3b-4a = 0 \qquad \cdots(2)$
Rearrange equation $(2)$ to get $b$ on its own. Add $4a$ to both sides, then divide both sides by $3$ .
$b = \frac{4a}{3}$
Put this in place of $b$ in equation $(1)$ , so only $a$ is left.
$3a+4\left(\frac{4a}{3}\right) = 1$
Work out the second term. Multiplying $\frac{4a}{3}$ by $4$ gives $\frac{16a}{3}$ .
$3a+\frac{16a}{3} = 1$
Multiply every term by $3$ to clear the fraction.
$9a+16a = 3$
Add the two terms on the left.
$25a = 3$
Divide both sides by $25$ .
$a = \frac{3}{25}$
Put this value back into $b=\frac{4a}{3}$ .
$b = \frac{4}{3}\times\frac{3}{25}$
The two $3$ s cancel.
$b = \frac{4}{25}$
State both answers.
$a = \frac{3}{25}, \qquad b = \frac{4}{25}$

Answer

a=\dfrac{3}{25}

and

b=\dfrac{4}{25}

Compact solution

Multiply the left side out term by term. Both $a$ and $b$ are real.

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

Equal complex numbers match part for part, and the right side is $1+0i$ .

\begin{aligned}3a+4b &= 1 \qquad\cdots(1) \\ 3b-4a &= 0 \qquad\cdots(2)\end{aligned}

Make $b$ the subject of $(2)$ .

\begin{aligned}3b &= 4a \\ b &= \frac{4a}{3}\end{aligned}

Put that into $(1)$ , so only $a$ is left.

\begin{aligned}3a+4\left(\frac{4a}{3}\right) &= 1 \\ 3a+\frac{16a}{3} &= 1 \\ 9a+16a &= 3 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{multiply every term by $3$}$}} \\ 25a &= 3 \\ a &= \frac{3}{25}\end{aligned}

Put that value back into $b=\frac{4a}{3}$ .

\begin{aligned}b &= \frac{4}{3}\times\frac{3}{25} \\ &= \frac{4}{25} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{the two $3$s cancel}$}}\end{aligned}

Thus the solution is

\begin{aligned}a &= \frac{3}{25}, \qquad b = \frac{4}{25}\end{aligned}

Check · Alternative method

Check the answer

\begin{aligned}(3-4i)\left(\frac{3}{25}+\frac{4}{25}i\right) &= \frac{(3-4i)(3+4i)}{25} \\ &= \frac{9+16}{25} \\ &= 1 \quad \checkmark\end{aligned}

Put the two values back in. The brackets give

a^{2}-b^{2}

with

a=3

and

b=4i

, which comes to

25

, so the answer is

1

Alternative methodThe question is really asking for the multiplicative inverse of

3-4i

, because

a+bi

is what

3-4i

must be multiplied by to give

1

Question 10

Solve the equation for $x$ and $y$ : $\ (3-2i)(x+yi)=2(x-2yi)+2i-1$ . Assume $x$ and $y$ are real.

Solution

Compact solution

Neither side is split into a real part and an imaginary part yet, so open the brackets on both sides. Take the left side first, multiplying term by term. Here $x$ and $y$ are real numbers.
$(3-2i)(x+yi) = 3x+3yi-2xi-2yi^{2}$
Replace $i^{2}$ by $-1$ . Note that $-2yi^{2}=2y$ , because minus times minus gives plus.
$(3-2i)(x+yi) = 3x+3yi-2xi+2y$
Collect the terms with no $i$ and the terms with $i$ , taking $i$ out as a common factor.
$(3-2i)(x+yi) = (3x+2y)+(3y-2x)i$
Now the right side. Open the bracket, multiplying each term inside by $2$ .
$2(x-2yi)+2i-1 = 2x-4yi+2i-1$
Collect it the same way. The terms with no $i$ are $2x$ and $-1$ , and the terms with $i$ are $-4yi$ and $2i$ .
$2(x-2yi)+2i-1 = (2x-1)+(2-4y)i$
Two complex numbers are equal only when both parts match. Compare the real parts first.
$3x+2y = 2x-1$
Take $2x$ from both sides to gather the $x$ terms, and number the result.
$x+2y = -1 \qquad \cdots(1)$
Now compare the imaginary parts.
$3y-2x = 2-4y$
Add $4y$ to both sides to gather the $y$ terms, and number the result.
$-2x+7y = 2 \qquad \cdots(2)$
Rearrange equation $(1)$ to get $x$ on its own, by taking $2y$ from both sides. We choose $x$ because it has no number in front, which keeps the working simple.
$x = -1-2y$
Put this in place of $x$ in equation $(2)$ , so only $y$ is left.
$-2(-1-2y)+7y = 2$
Open the bracket. Minus times minus gives plus, so $-2\times(-1)=2$ and $-2\times(-2y)=4y$ .
$2+4y+7y = 2$
Add the two $y$ terms.
$2+11y = 2$
Take $2$ from both sides.
$11y = 0$
Divide both sides by $11$ . $0$ shared between any number of parts is still $0$ .
$y = 0$
Put $y=0$ back into $x=-1-2y$ . Twice $0$ is $0$ , so nothing is taken away.
$x = -1-2(0) = -1$
State both answers.
$x = -1, \qquad y = 0$

Answer

x=-1

and

y=0

Compact solution

Open the left side term by term. Both $x$ and $y$ are real.

\begin{aligned}(3-2i)(x+yi) &= 3x+3yi-2xi-2yi^{2} \\ &= 3x+3yi-2xi-2y(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^{2}=-1$}} \\ &= 3x+3yi-2xi+2y \\ &= (3x+2y)+(3y-2x)i\end{aligned}

Now the right side, opened and collected the same way.

\begin{aligned}2(x-2yi)+2i-1 &= 2x-4yi+2i-1 \\ &= (2x-1)+(2-4y)i\end{aligned}

Equal complex numbers match part for part. Compare the real parts.

\begin{aligned}3x+2y &= 2x-1 \\ x+2y &= -1 \qquad\cdots(1)\end{aligned}

Then compare the imaginary parts.

\begin{aligned}3y-2x &= 2-4y \\ -2x+7y &= 2 \qquad\cdots(2)\end{aligned}

In $(1)$ the letter $x$ carries no number in front, so make it the subject.

\begin{aligned}x &= -1-2y\end{aligned}

Put that into $(2)$ , so only $y$ is left.

\begin{aligned}-2(-1-2y)+7y &= 2 \\ 2+4y+7y &= 2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$-2(-1-2y)=2+4y$}} \\ 2+11y &= 2 \\ 11y &= 0 \\ y &= 0\end{aligned}

Put that back into $x=-1-2y$ .

\begin{aligned}x &= -1-2(0) \\ &= -1\end{aligned}

Thus the solution is

\begin{aligned}x &= -1, \qquad y = 0\end{aligned}

Check · Common mistake

Check the answer

\begin{aligned}(3-2i)(-1+0i) &= -3+2i \\ 2(-1-0)+2i-1 &= -3+2i \quad \checkmark\end{aligned}

Put

x=-1

and

y=0

into each side on its own. Both come to

-3+2i

, so the answers are correct.

Watch outThe value

y=0

is a valid solution, not an omitted answer. It says the number

x+yi

turns out to be the real number

-1

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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

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

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