Class 10 Exercise 1.4 Solutions — Complex Numbers

Concepts

Rule

Real and imaginary parts of a power

Let $z=(x+iy)^{n}$ , with $x,y$ real and $n\in\{1,-1,2,-2\}$ . If $n<0$ , assume $x+iy\neq0$ . Rewrite $z$ as $u+iv$ , with $u,v$ real. Then $\operatorname{Re}(z)=u,\qquad\operatorname{Im}(z)=v$ The real and imaginary parts exist before the rewriting; the calculation makes their values explicit.

The $i$ is never part of the imaginary part. For $z=55-48i$ , $\operatorname{Im}(z)=-48$ , not $-48i$ .

For a nonzero base and a positive integer $n$ , $(x+iy)^{-n}=1/(x+iy)^{n}$ . A negative exponent means a reciprocal, not a minus sign in front of the answer.

For $n=-1$ the imaginary part carries a minus sign, because the conjugate on top is $x-iy$ .

Examples

For $z=(2+i)^{1}$ : $\ \operatorname{Re}(z)=2$ and $\operatorname{Im}(z)=1$ .
For $z=(4+3i)^{-1}=\frac{4-3i}{25}$ : $\ \operatorname{Re}(z)=\frac{4}{25}$ and $\operatorname{Im}(z)=-\frac{3}{25}$ .
For $z=(4+3i)^{-2}=\frac{7-24i}{625}$ : $\ \operatorname{Re}(z)=\frac{7}{625}$ and $\operatorname{Im}(z)=-\frac{24}{625}$ .

Definition

Powers of a quotient

To identify the parts of a power of a quotient, rewrite the result in rectangular form. You may simplify the quotient first. For a negative power, taking the reciprocal first is often shorter; then simplify that quotient and apply the remaining positive power. The original denominator must be nonzero, and for a negative power the original numerator must also be nonzero.

If $p\neq0$ and $q\neq0$ , then $\left(\frac pq\right)^{-1}=\frac qp$ . Both restrictions matter: the original quotient and its reciprocal must be defined.

A power of $-2$ turns the fraction upside down and then squares it.

Once the number is in the form $x+iy$ , the real part is $x$ and the imaginary part is $y$ .

Examples

For $z=\left(\frac{4+3i}{3+2i}\right)^{-1}=\frac{18-i}{25}$ , we have $\operatorname{Re}(z)=\frac{18}{25}$ and $\operatorname{Im}(z)=-\frac1{25}$ .
For $z=\left(\frac{1+i}{1-i}\right)^{-2}=-1$ , we have $\operatorname{Re}(z)=-1$ and $\operatorname{Im}(z)=0$ .

Definition

Real and imaginary parts from the conjugate

The two parts can also be read from the number and its conjugate: $\operatorname{Re}(z)=\frac{z+\bar{z}}{2}, \qquad \operatorname{Im}(z)=\frac{z-\bar{z}}{2i}$

Adding $z$ and $\bar{z}$ cancels the imaginary parts and doubles the real part. Subtracting them cancels the real parts and leaves twice the imaginary term.

This is mostly useful as a check. Simplifying the number to the form $x+iy$ is usually the quicker route.

Examples

For $z=3-5i$ : $\ z+\bar{z}=(3-5i)+(3+5i)=6$ , so $\operatorname{Re}(z)=\frac{6}{2}=3$ .
For $z=3-5i$ : $\ z-\bar{z}=(3-5i)-(3+5i)=-10i$ , so $\operatorname{Im}(z)=\frac{-10i}{2i}=-5$ .

Definition

Simultaneous linear equations with complex coefficients

A system of simultaneous equations asks for values of the unknowns that satisfy every equation in the system. Linear systems with complex coefficients can be solved by substitution or elimination, using the same algebraic laws as for real coefficients.

In substitution, express one unknown in terms of the other, substitute, solve the resulting equation, and substitute back. In elimination, add or subtract suitable multiples of the equations to remove one unknown.

The working is the same as for real equations. The one extra move is clearing $i$ from a denominator with the conjugate whenever a division appears.

Choose the equation and the unknown that are simplest to isolate, usually one with coefficient $1$ or $-1$ .

Put both answers back into both original equations to check.

Examples

The system $5z-(3+i)w=7-i$ and $(2-i)z+2iw=4$ has the solution $z=2$ and $w=1$ .
The system $z-4w=3i$ and $2z+3w=11-5i$ has the solution $z=4-i$ and $w=1-i$ .

Deep understanding

Reading the parts after simplification

The real part and the imaginary part are names for the two pieces of a number written as $x+iy$ : one plain real number, then a real number times $i$ .

The expressions $(8-3i)^{2}$ and $(4+3i)^{-1}$ have real and imaginary parts, but these are not simply the coordinates written inside the original brackets.

Multiplying the bracket out, or clearing the $i$ from a denominator, does not change the number. It only rewrites it in the standard shape. $\begin{aligned}(8-3i)^{2} &= 8^{2}-2(8)(3i)+(3i)^{2} \\ &= 64-48i+9i^{2} \\ &= 64-48i-9 \\ &= 55-48i\end{aligned}$

Now that it reads $55-48i$ , the real part is $55$ and the imaginary part is $-48$ . These parts existed throughout; rewriting makes them easy to read.

Why $(x+iy)^{-1}$ has the parts it does

Assume $x,y$ are real and $x+iy\neq0$ . Start from the meaning of a negative exponent: $\begin{aligned}z&=(x+iy)^{-1}=\frac{1}{x+iy}\end{aligned}$

There is an $i$ in the denominator, so this is not the form $x+iy$ . Multiply the top and the bottom by the conjugate $x-iy$ . You may do this because $\frac{x-iy}{x-iy}=1$ , and multiplying by $1$ changes how a number looks but not its value. $\begin{aligned}z &= \frac{1}{x+iy}\times\frac{x-iy}{x-iy}\end{aligned}$

The denominator is of the form $(x+iy)(x-iy)$ , which is $x^{2}-(iy)^{2}=x^{2}+y^{2}$ , an ordinary real number with no $i$ in it. $\begin{aligned}z &= \frac{x-iy}{x^{2}+y^{2}}\end{aligned}$

Split the one fraction into two, one for each part. $\begin{aligned}z &= \frac{x}{x^{2}+y^{2}} - \frac{y}{x^{2}+y^{2}}\,i\end{aligned}$

So $\operatorname{Re}(z)=\frac{x}{x^{2}+y^{2}}$ and $\operatorname{Im}(z)=\frac{-y}{x^{2}+y^{2}}$ . The minus sign sits on the imaginary part, because the conjugate on top was $x-iy$ .

The parts of a square and a reciprocal square

Let $a=x+iy$ , where $x,y$ are real. Expanding the square gives

$\begin{aligned}a^{2}&=(x+iy)^{2} \\ &=x^{2}+2xyi+y^{2}i^{2} \\ &=(x^{2}-y^{2})+2xyi\end{aligned}$

Thus $\operatorname{Re}(a^{2})=x^{2}-y^{2}$ and $\operatorname{Im}(a^{2})=2xy$ .

For $a\neq0$ , use $a^{-1}=(x-iy)/(x^{2}+y^{2})$ and square both the numerator and the denominator:

$\begin{aligned}a^{-2}&=\left(\frac{x-iy}{x^{2}+y^{2}}\right)^{2} \\ &=\frac{x^{2}-2xyi+y^{2}i^{2}}{(x^{2}+y^{2})^{2}} \\ &=\frac{x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}-i\frac{2xy}{(x^{2}+y^{2})^{2}}\end{aligned}$

The denominator is positive because $x^{2}+y^{2}>0$ . Therefore,

$\begin{aligned}\operatorname{Re}(a^{-2})&=\frac{x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}, \\ \operatorname{Im}(a^{-2})&=-\frac{2xy}{(x^{2}+y^{2})^{2}}\end{aligned}$

Why substitution solves a system with complex coefficients

Two equations may determine two unknowns, but the equations must provide independent, consistent information. They can also have no solution or infinitely many solutions.

Making $w$ the subject of one equation turns the other equation into a single equation whose only unknown is $z$ .

In the examples here, elimination leaves a nonzero coefficient of the remaining unknown, so division gives one value. If the coefficient were $0$ , we would first check whether the resulting equation is an identity or a contradiction.

With $z$ known, the rearranged first equation hands you $w$ straight away.

For example, from $5z-(3+i)w=7-i$ we get $w=\frac{5z-7+i}{3+i}$ . Putting this into $(2-i)z+2iw=4$ and clearing the fraction leads to $z(7+9i)=2(7+9i)$ , so $z=2$ , and then $w=\frac{3+i}{3+i}=1$ .

Recovering the two parts from conjugates

If $z=x+iy$ , then $\bar z=x-iy$ . Addition cancels the imaginary terms, while subtraction cancels the real terms:

$\begin{aligned}z+\bar z&=(x+iy)+(x-iy)=2x, \\ z-\bar z&=(x+iy)-(x-iy)=2iy\end{aligned}$

Divide the first equation by $2$ and the second by $2i$ . Both divisors are nonzero, so

$\begin{aligned}\operatorname{Re}(z)&=\frac{z+\bar z}{2}, \\ \operatorname{Im}(z)&=\frac{z-\bar z}{2i}\end{aligned}$

Examples

Example 17

Find the real and imaginary parts of $(4+3i)^{-1}$ .

Solution

Compact solution

Call the number $z$ . A power of $-1$ means turn the number over, using $a^{-n}=\frac{1}{a^{n}}$ , so write it as $1$ over the bracket.
$z = (4+3i)^{-1} = \frac{1}{4+3i}$
There is an $i$ in the denominator, so this is not yet the form $x+iy$ . Multiply the top and the bottom by the conjugate of the bottom, which is $4-3i$ .
$z = \frac{1}{4+3i}\times\frac{4-3i}{4-3i}$
Work out the denominator. The two brackets are 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. For the second, square both the $3$ and the $i$ , and $i^{2}=-1$ , so $(3i)^{2}=-9$ .
$(4+3i)(4-3i) = 16-(-9) = 25$
The numerator is just $4-3i$ , because the top was $1$ . Put it over the denominator.
$z = \frac{4-3i}{25}$
Split it into a real part and an imaginary part.
$z = \frac{4}{25}-\frac{3}{25}i$
The real and imaginary parts are
$\operatorname{Re}(z) = \frac{4}{25}, \qquad \operatorname{Im}(z) = -\frac{3}{25}$

Answer

\mathrm{Re}(z)=\dfrac{4}{25}

and

\mathrm{Im}(z)=-\dfrac{3}{25}

Compact solution

Let $z=(4+3i)^{-1}$ . Write the reciprocal and use the conjugate $4-3i$ .

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

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

\begin{gathered}\operatorname{Re}(z)=\dfrac{4}{25} \\ \operatorname{Im}(z)=-\dfrac{3}{25}\end{gathered}

Example 18

Find the real and imaginary parts of $z=(4+3i)^{-2}$ .

Solution

Compact solution

The power is $-2$ . Turn the number over first, so the square moves to the denominator.
$z = (4+3i)^{-2} = \frac{1}{(4+3i)^{2}}$
Square the bracket. Use the identity $(a+b)^{2}=a^{2}+2ab+b^{2}$ with $a=4$ and $b=3i$ .
$(4+3i)^{2} = 4^{2}+2(4)(3i)+(3i)^{2}$
Work out the three pieces. The middle term is $24i$ , and $(3i)^{2}=9i^{2}$ .
$(4+3i)^{2} = 16+24i+9i^{2}$
Replace $i^{2}$ by $-1$ , so $9i^{2}=-9$ .
$(4+3i)^{2} = 16+24i-9$
Collect the two real numbers. Taking $9$ from $16$ leaves $7$ .
$(4+3i)^{2} = 7+24i$
Put that back underneath.
$z = \frac{1}{7+24i}$
There is an $i$ in the denominator, so multiply the top and the bottom by the conjugate $7-24i$ .
$z = \frac{1}{7+24i}\times\frac{7-24i}{7-24i}$
The denominator is of the form $(a+b)(a-b)=a^{2}-b^{2}$ , and $(24i)^{2}=576i^{2}=-576$ .
$(7+24i)(7-24i) = 49-(-576) = 625$
The numerator is $7-24i$ . Put it over the denominator.
$z = \frac{7-24i}{625}$
Split it into a real part and an imaginary part.
$z = \frac{7}{625}-\frac{24}{625}i$
The real and imaginary parts are
$\operatorname{Re}(z) = \frac{7}{625}, \qquad \operatorname{Im}(z) = -\frac{24}{625}$

Answer

\mathrm{Re}(z)=\dfrac{7}{625}

and

\mathrm{Im}(z)=-\dfrac{24}{625}

Compact solution

Let $z=(4+3i)^{-2}$ . Rewrite the negative exponent as a reciprocal.

z=\frac{1}{(4+3i)^2}

Expand the square, keeping the middle term.

\begin{aligned}(4+3i)^2 &= 4^2+2(4)(3i)+(3i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)^2=a^2+2ab+b^2$}} \\ &= 16+24i+9i^2 \\ &= 16+24i+9(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= (16-9)+24i \\ &= 7+24i\end{aligned}

Substitute the square into the denominator. Write the reciprocal and use the conjugate $7-24i$ .

\begin{aligned}z &= \frac{1}{7+24i} \\ &= \frac{1}{7+24i}\cdot\frac{7-24i}{7-24i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{7-24i}{7-24i}=1$}} \\ &= \frac{7-24i}{(7+24i)(7-24i)}\end{aligned}

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(7+24i)(7-24i) &= 7^2-(24i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 49-576i^2 \\ &= 49-576(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 49+576 \\ &= 625\end{aligned}

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

\begin{aligned}z &= \frac{7-24i}{625} \\ &= \frac{7}{625}-\frac{24}{625}i\end{aligned}

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

\begin{gathered}\operatorname{Re}(z)=\dfrac{7}{625} \\ \operatorname{Im}(z)=-\dfrac{24}{625}\end{gathered}

Common mistake

Watch outThe exponent

-2

means the reciprocal of a square, not the negative of a square. Squaring first and taking the reciprocal first are both valid here because the base is nonzero.

Example 19

Find the real and imaginary parts of $z=\left(\frac{4+3i}{3+2i}\right)^{-1}$ .

Solution

Compact solution

A power of $-1$ on a fraction turns the fraction upside down. This is the rule $\left(\frac{p}{q}\right)^{-1}=\frac{q}{p}$ .
$z = \left(\frac{4+3i}{3+2i}\right)^{-1} = \frac{3+2i}{4+3i}$
There is an $i$ in the denominator, so multiply the top and the bottom by the conjugate of the bottom, which is $4-3i$ .
$z = \frac{3+2i}{4+3i}\times\frac{4-3i}{4-3i}$
Work out the numerator, multiplying term by term.
$(3+2i)(4-3i) = 12-9i+8i-6i^{2}$
Replace $i^{2}$ by $-1$ . Note that $-6i^{2}=6$ , because minus times minus gives plus.
$(3+2i)(4-3i) = 12-9i+8i+6$
Collect. The real numbers give $12+6=18$ , and the $i$ terms give $-9i+8i=-i$ .
$(3+2i)(4-3i) = 18-i$
Work out the denominator, of the form $(a+b)(a-b)=a^{2}-b^{2}$ .
$(4+3i)(4-3i) = 4^{2}-(3i)^{2} = 16+9 = 25$
Put the numerator over the denominator.
$z = \frac{18-i}{25}$
Split it into a real part and an imaginary part.
$z = \frac{18}{25}-\frac{1}{25}i$
The real and imaginary parts are
$\operatorname{Re}(z) = \frac{18}{25}, \qquad \operatorname{Im}(z) = -\frac{1}{25}$

Answer

\mathrm{Re}(z)=\dfrac{18}{25}

and

\mathrm{Im}(z)=-\dfrac{1}{25}

Compact solution

Let $z=\left(\frac{4+3i}{3+2i}\right)^{-1}$ . Take the reciprocal of the nonzero fraction, retaining the positive exponent.

z=\left(\frac{3+2i}{4+3i}\right)^{1}

Multiply numerator and denominator by the conjugate $4-3i$ .

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

Expand the numerator first.

\begin{aligned}(3+2i)(4-3i) &= 3(4)+3(-3i)+(2i)(4)+(2i)(-3i) \\ &= 12-9i+8i-6i^2 \\ &= 12-9i+8i-6(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 12-9i+8i+6 \\ &= (12+6)+(-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}(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}\frac{3+2i}{4+3i} &= \frac{18-i}{25} \\ &= \frac{18}{25}-\frac{1}{25}i\end{aligned}

The simplified quotient is

z=\dfrac{18}{25}-\dfrac{1}{25}i

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

\begin{gathered}\operatorname{Re}(z)=\dfrac{18}{25} \\ \operatorname{Im}(z)=-\dfrac{1}{25}\end{gathered}

Example 20

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

Solution

Compact solution

A power of $-2$ turns the fraction upside down and then squares it. Flip it first.
$z = \left(\frac{1+i}{1-i}\right)^{-2} = \left(\frac{1-i}{1+i}\right)^{2}$
Simplify the fraction inside to the form $x+iy$ before squaring. Multiply the top and the bottom by the conjugate of the bottom, which is $1-i$ .
$\frac{1-i}{1+i} = \frac{1-i}{1+i}\times\frac{1-i}{1-i}$
The top is now $(1-i)^{2}$ . Use $(a-b)^{2}=a^{2}-2ab+b^{2}$ with $a=1$ and $b=i$ .
$(1-i)^{2} = 1^{2}-2(1)(i)+i^{2}$
Work it out, with $i^{2}=-1$ .
$(1-i)^{2} = 1-2i-1 = -2i$
The bottom is $(1+i)(1-i)$ , of the form $(a+b)(a-b)=a^{2}-b^{2}$ .
$(1+i)(1-i) = 1^{2}-i^{2} = 1+1 = 2$
So the fraction inside becomes this.
$\frac{1-i}{1+i} = \frac{-2i}{2} = -i$
Now square it. The square of $-i$ is $(-1)^{2}i^{2}=i^{2}$ .
$z = (-i)^{2} = i^{2}$
Replace $i^{2}$ by $-1$ . Written in full complex form this is $-1+0i$ .
$z = -1 = -1+0i$
The real and imaginary parts are
$\operatorname{Re}(z) = -1, \qquad \operatorname{Im}(z) = 0$

Answer

\mathrm{Re}(z)=-1

and

\mathrm{Im}(z)=0

Compact solution

Let $z=\left(\frac{1+i}{1-i}\right)^{-2}$ . Take the reciprocal of the nonzero fraction, retaining the positive exponent.

z=\left(\frac{1-i}{1+i}\right)^{2}

Multiply numerator and denominator by the conjugate $1-i$ .

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

Expand the numerator first.

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

Now square the simplified quotient. Square the real coefficient and the factor $i$ separately.

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

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

\begin{gathered}\operatorname{Re}(z)=-1 \\ \operatorname{Im}(z)=0\end{gathered}

Common mistake

Watch outThe answer is a real number. An expression built from

i

does not have to stay imaginary.

Example 21

Solve the simultaneous linear equations $5z-(3+i)w=7-i$ and $(2-i)z+2iw=4$ for $z$ and $w$ .

Solution

The given equations are

\begin{aligned}5z-(3+i)w&=7-i,\qquad\cdots(1) \\ (2-i)z+2iw&=4.\qquad\cdots(2)\end{aligned}

Compact solution

Call the two equations $(1)$ and $(2)$ . Solve equation $(1)$ for $w$ in terms of $z$ , by moving the $z$ term across and then dividing.
$5z-(3+i)w = 7-i \qquad \cdots(1)$
Add $(3+i)w$ to both sides and take $7-i$ from both sides, so the $w$ term is by itself.
$(3+i)w = 5z-7+i$
Divide both sides by $3+i$ to make $w$ the subject. Call this equation $(3)$ .
$w = \frac{5z-7+i}{3+i} \qquad \cdots(3)$
Put this expression for $w$ into equation $(2)$ .
$(2-i)z+2i\left(\frac{5z-7+i}{3+i}\right) = 4$
Multiply every term by $3+i$ to clear the fraction.
$(2-i)(3+i)z+2i(5z-7+i) = 4(3+i)$
Work out $(2-i)(3+i)$ term by term.
$(2-i)(3+i) = 6+2i-3i-i^{2}$
Replace $i^{2}$ by $-1$ and collect. $-i^{2}=1$ , so $6+1=7$ , and $2i-3i=-i$ .
$(2-i)(3+i) = 7-i$
Work out $2i(5z-7+i)$ . The last term is $2i^{2}=-2$ .
$2i(5z-7+i) = 10iz-14i-2$
Put these back. The right side is $4(3+i)=12+4i$ .
$(7-i)z+10iz-14i-2 = 12+4i$
Add $14i+2$ to both sides, so the two $z$ terms stand alone on the left.
$(7-i)z+10iz = 14+18i$
Add the two $z$ terms. $(7-i)+10i=7+9i$ .
$(7+9i)z = 14+18i$
The right side is $2(7+9i)$ , so divide both sides by $7+9i$ .
$z = \frac{2(7+9i)}{7+9i} = 2$
Put $z=2$ into equation $(3)$ .
$w = \frac{5(2)-7+i}{3+i}$
Work out the top. $10-7+i=3+i$ , which is the same as the bottom.
$w = \frac{3+i}{3+i} = 1$
State both answers.
$z = 2, \qquad w = 1$

Answer

z=2

and

w=1

Compact solution

Label the two original equations.

\begin{aligned}5z-(3+i)w&=7-i\qquad\cdots(1) \\ (2-i)z+2iw&=4\qquad\cdots(2)\end{aligned}

Use equation $(1)$ to make $w$ the subject.

\begin{aligned}5z-(3+i)w &= 7-i \qquad\cdots(1) \\ (3+i)w &= 5z-7+i \\ w &= \frac{5z-7+i}{3+i} \qquad\cdots(3)\end{aligned}

Put that into equation $(2)$ , then multiply every term by $3+i$ to clear the fraction.

\begin{aligned}(2-i)z+2i\left(\frac{5z-7+i}{3+i}\right) &= 4 \\ (2-i)(3+i)z+2i(5z-7+i) &= 4(3+i)\end{aligned}

First expand $(2-i)(3+i)$ .

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

Next expand $2i(5z-7+i)$ .

\begin{aligned}2i(5z-7+i) &= 10iz-14i+2i^{2} \\ &= 10iz-14i-2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^{2}=-1$}}\end{aligned}

Put both back, with $4(3+i)=12+4i$ on the right.

\begin{aligned}(7-i)z+10iz-14i-2 &= 12+4i \\ (7-i)z+10iz &= 14+18i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{add $14i+2$ to both sides}$}} \\ (7+9i)z &= 14+18i \\ (7+9i)z &= 2(7+9i) \\ z &= 2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$7+9i\neq0$}}\end{aligned}

Put $z=2$ back into equation $(3)$ .

\begin{aligned}w &= \frac{5(2)-7+i}{3+i} \\ &= \frac{3+i}{3+i} \\ &= 1\end{aligned}

Thus the solution is

\begin{aligned}z &= 2, \qquad w = 1\end{aligned}

Check

Check the answer

\begin{aligned}5(2)-(3+i)(1)&=10-3-i=7-i\\(2-i)(2)+2i(1)&=4-2i+2i=4\end{aligned}

Substitute the stated pair into both original equations. Each left side gives its own right side, so the pair satisfies the entire system.

Exercise 1.4

Question 1

Find the real and imaginary parts of the following complex numbers:

(i)

(8-3i)^{2}

Solution

Compact solution

Let $z$ denote the given expression. Squaring a bracket uses the identity $(a-b)^{2}=a^{2}-2ab+b^{2}$ . Here $a=8$ and $b=3i$ , so write that out.
$(8-3i)^{2} = 8^{2}-2(8)(3i)+(3i)^{2}$
Work out each piece. $8^{2}=64$ , the middle term is $-48i$ , and $(3i)^{2}=9i^{2}$ .
$(8-3i)^{2} = 64-48i+9i^{2}$
Replace $i^{2}$ by $-1$ , so $9i^{2}=-9$ .
$(8-3i)^{2} = 64-48i-9$
Collect the two real numbers. Taking $9$ from $64$ leaves $55$ .
$(8-3i)^{2} = 55-48i$
The number is now in the form $x+iy$ . The real part is the piece with no $i$ , and the imaginary part is the number multiplying $i$ .
$\operatorname{Re}(z) = 55, \qquad \operatorname{Im}(z) = -48$

Answer

\mathrm{Re}(z)=55

and

\mathrm{Im}(z)=-48

Compact solution

Let $z=(8-3i)^{2}$ . Expand the square, keeping the middle term.

\begin{aligned}z &= (8-3i)^2 \\ &= 8^2-2(8)(3i)+(3i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a-b)^2=a^2-2ab+b^2$}} \\ &= 64-48i+9i^2 \\ &= 64-48i+9(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= (64-9)-48i \\ &= 55-48i\end{aligned}

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

\begin{gathered}\operatorname{Re}(z)=55 \\ \operatorname{Im}(z)=-48\end{gathered}

(ii)

(5+3i)^{-1}

Solution

Compact solution

Let $z$ denote the given expression. A power of $-1$ means turn the number over, using $a^{-n}=\frac{1}{a^{n}}$ . Write it as $1$ over the bracket.
$(5+3i)^{-1} = \frac{1}{5+3i}$
There is an $i$ in the denominator, so this is not the form $x+iy$ yet. Multiply the top and the bottom by the conjugate of the bottom, which is $5-3i$ .
$(5+3i)^{-1} = \frac{1}{5+3i}\times\frac{5-3i}{5-3i}$
Work out the denominator. The two brackets are of the form $(a+b)(a-b)$ , which is always $a^{2}-b^{2}$ .
$(5+3i)(5-3i) = 5^{2}-(3i)^{2}$
Square each part. For the second, square both the $3$ and the $i$ , and $i^{2}=-1$ , so $(3i)^{2}=-9$ .
$(5+3i)(5-3i) = 25-(-9) = 34$
The numerator is just $5-3i$ , because the top was $1$ . Put it over the denominator.
$(5+3i)^{-1} = \frac{5-3i}{34}$
Split it into a real part and an imaginary part.
$(5+3i)^{-1} = \frac{5}{34}-\frac{3}{34}i$
The real and imaginary parts are
$\operatorname{Re}(z) = \frac{5}{34}, \qquad \operatorname{Im}(z) = -\frac{3}{34}$

Answer

\mathrm{Re}(z)=\dfrac{5}{34}

and

\mathrm{Im}(z)=-\dfrac{3}{34}

Compact solution

Let $z=(5+3i)^{-1}$ . Write the reciprocal and use the conjugate $5-3i$ .

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

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(5+3i)(5-3i) &= 5^2-(3i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 25-9i^2 \\ &= 25-9(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 25+9 \\ &= 34\end{aligned}

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

\begin{aligned}z &= \frac{5-3i}{34} \\ &= \frac{5}{34}-\frac{3}{34}i\end{aligned}

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

\begin{gathered}\operatorname{Re}(z)=\dfrac{5}{34} \\ \operatorname{Im}(z)=-\dfrac{3}{34}\end{gathered}

(iii)

(4-5i)^{-1}

Solution

Compact solution

Let $z$ denote the given expression. Here the power is $-1$ , so write the number as $1$ over the bracket.
$(4-5i)^{-1} = \frac{1}{4-5i}$
Clear the $i$ from the denominator by multiplying the top and the bottom by the conjugate of the bottom, which is $4+5i$ .
$(4-5i)^{-1} = \frac{1}{4-5i}\times\frac{4+5i}{4+5i}$
Work out the denominator, of the form $(a-b)(a+b)=a^{2}-b^{2}$ .
$(4-5i)(4+5i) = 4^{2}-(5i)^{2}$
Square each part. Since $(5i)^{2}=25i^{2}=-25$ , taking it away adds $25$ .
$(4-5i)(4+5i) = 16-(-25) = 41$
The numerator is $4+5i$ , because the top was $1$ .
$(4-5i)^{-1} = \frac{4+5i}{41}$
Split it into a real part and an imaginary part. Both parts are positive here.
$(4-5i)^{-1} = \frac{4}{41}+\frac{5}{41}i$
The real and imaginary parts are
$\operatorname{Re}(z) = \frac{4}{41}, \qquad \operatorname{Im}(z) = \frac{5}{41}$

Answer

\mathrm{Re}(z)=\dfrac{4}{41}

and

\mathrm{Im}(z)=\dfrac{5}{41}

Compact solution

Let $z=(4-5i)^{-1}$ . Write the reciprocal and use the conjugate $4+5i$ .

\begin{aligned}z &= \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 &= \frac{4+5i}{41} \\ &= \frac{4}{41}+\frac{5}{41}i\end{aligned}

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

\begin{gathered}\operatorname{Re}(z)=\dfrac{4}{41} \\ \operatorname{Im}(z)=\dfrac{5}{41}\end{gathered}

Common mistake

Watch outThe imaginary part is

+\frac{5}{41}

, not

-\frac{5}{41}

. The conjugate of

4-5i

4+5i

, so the

i

term on top is

+5i

(iv)

(4-3i)^{-2}

Solution

Compact solution

The exponent is $-2$ , so take the reciprocal and leave the positive square in the denominator.
$(4-3i)^{-2} = \frac{1}{(4-3i)^{2}}$
Square the bracket. Use the identity $(a-b)^{2}=a^{2}-2ab+b^{2}$ with $a=4$ and $b=3i$ .
$(4-3i)^{2} = 4^{2}-2(4)(3i)+(3i)^{2}$
Work out the three pieces. The middle term is $-24i$ , and $(3i)^{2}=9i^{2}$ .
$(4-3i)^{2} = 16-24i+9i^{2}$
Replace $i^{2}$ by $-1$ , so $9i^{2}=-9$ .
$(4-3i)^{2} = 16-24i-9$
Collect the real numbers. Taking $9$ from $16$ leaves $7$ .
$(4-3i)^{2} = 7-24i$
Put that back underneath.
$(4-3i)^{-2} = \frac{1}{7-24i}$
Clear the $i$ from the denominator by multiplying the top and the bottom by the conjugate $7+24i$ .
$(4-3i)^{-2} = \frac{1}{7-24i}\times\frac{7+24i}{7+24i}$
The denominator is of the form $(a-b)(a+b)=a^{2}-b^{2}$ , and $(24i)^{2}=576i^{2}=-576$ .
$(7-24i)(7+24i) = 49-(-576) = 625$
The numerator is $7+24i$ .
$(4-3i)^{-2} = \frac{7+24i}{625}$
Split it into a real part and an imaginary part.
$(4-3i)^{-2} = \frac{7}{625}+\frac{24}{625}i$
The real and imaginary parts are
$\operatorname{Re}(z) = \frac{7}{625}, \qquad \operatorname{Im}(z) = \frac{24}{625}$

Answer

\mathrm{Re}(z)=\dfrac{7}{625}

and

\mathrm{Im}(z)=\dfrac{24}{625}

Compact solution

Let $z=(4-3i)^{-2}$ . Rewrite the negative exponent as a reciprocal.

z=\frac{1}{(4-3i)^2}

Expand the square, keeping the middle term.

\begin{aligned}(4-3i)^2 &= 4^2-2(4)(3i)+(3i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a-b)^2=a^2-2ab+b^2$}} \\ &= 16-24i+9i^2 \\ &= 16-24i+9(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= (16-9)-24i \\ &= 7-24i\end{aligned}

Substitute the square into the denominator. Write the reciprocal and use the conjugate $7+24i$ .

\begin{aligned}z &= \frac{1}{7-24i} \\ &= \frac{1}{7-24i}\cdot\frac{7+24i}{7+24i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\frac{7+24i}{7+24i}=1$}} \\ &= \frac{7+24i}{(7-24i)(7+24i)}\end{aligned}

Multiply the conjugate pair using the difference of squares.

\begin{aligned}(7-24i)(7+24i) &= 7^2-(24i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^2-b^2$}} \\ &= 49-576i^2 \\ &= 49-576(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 49+576 \\ &= 625\end{aligned}

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

\begin{aligned}z &= \frac{7+24i}{625} \\ &= \frac{7}{625}+\frac{24}{625}i\end{aligned}

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

\begin{gathered}\operatorname{Re}(z)=\dfrac{7}{625} \\ \operatorname{Im}(z)=\dfrac{24}{625}\end{gathered}

(v)

\left(\frac{3+2i}{4+3i}\right)^{-1}

Solution

Compact solution

Let $z$ denote the given expression. A power of $-1$ on a fraction turns the fraction upside down. This is the rule $\left(\frac{p}{q}\right)^{-1}=\frac{q}{p}$ .
$\left(\frac{3+2i}{4+3i}\right)^{-1} = \frac{4+3i}{3+2i}$
There is an $i$ in the denominator, so multiply the top and the bottom by the conjugate of the bottom, which is $3-2i$ .
$\frac{4+3i}{3+2i} = \frac{4+3i}{3+2i}\times\frac{3-2i}{3-2i}$
Work out the numerator, multiplying term by term.
$(4+3i)(3-2i) = 12-8i+9i-6i^{2}$
Replace $i^{2}$ by $-1$ . Note that $-6i^{2}=6$ , because minus times minus gives plus.
$(4+3i)(3-2i) = 12-8i+9i+6$
Collect. The real numbers give $12+6=18$ , and the $i$ terms give $-8i+9i=i$ .
$(4+3i)(3-2i) = 18+i$
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.
$\left(\frac{3+2i}{4+3i}\right)^{-1} = \frac{18+i}{13}$
Split it into a real part and an imaginary part.
$\left(\frac{3+2i}{4+3i}\right)^{-1} = \frac{18}{13}+\frac{1}{13}i$
The real and imaginary parts are
$\operatorname{Re}(z) = \frac{18}{13}, \qquad \operatorname{Im}(z) = \frac{1}{13}$

Answer

\mathrm{Re}(z)=\dfrac{18}{13}

and

\mathrm{Im}(z)=\dfrac{1}{13}

Compact solution

Let $z=\left(\frac{3+2i}{4+3i}\right)^{-1}$ . Take the reciprocal of the nonzero fraction, retaining the positive exponent.

z=\left(\frac{4+3i}{3+2i}\right)^{1}

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

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

Expand the numerator first.

\begin{aligned}(4+3i)(3-2i) &= 4(3)+4(-2i)+(3i)(3)+(3i)(-2i) \\ &= 12-8i+9i-6i^2 \\ &= 12-8i+9i-6(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= 12-8i+9i+6 \\ &= (12+6)+(-8+9)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}(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{4+3i}{3+2i} &= \frac{18+i}{13} \\ &= \frac{18}{13}+\frac{1}{13}i\end{aligned}

The simplified quotient is

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

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

\begin{gathered}\operatorname{Re}(z)=\dfrac{18}{13} \\ \operatorname{Im}(z)=\dfrac{1}{13}\end{gathered}

(vi)

\left(\frac{2-i}{2+i}\right)^{-2}

Solution

Compact solution

Let $z$ denote the given expression. A power of $-2$ on a fraction turns the fraction upside down and then squares it. Flip it first.
$\left(\frac{2-i}{2+i}\right)^{-2} = \left(\frac{2+i}{2-i}\right)^{2}$
Simplify the fraction inside to the form $x+iy$ before squaring. Multiply the top and the bottom by the conjugate of the bottom, which is $2+i$ .
$\frac{2+i}{2-i} = \frac{2+i}{2-i}\times\frac{2+i}{2+i}$
The top is now $(2+i)^{2}$ . Use $(a+b)^{2}=a^{2}+2ab+b^{2}$ with $a=2$ and $b=i$ .
$(2+i)^{2} = 2^{2}+2(2)(i)+i^{2}$
Work it out, with $i^{2}=-1$ , so the last term is $-1$ .
$(2+i)^{2} = 4+4i-1 = 3+4i$
The bottom is $(2-i)(2+i)$ , of the form $(a-b)(a+b)=a^{2}-b^{2}$ .
$(2-i)(2+i) = 2^{2}-i^{2} = 4+1 = 5$
So the fraction inside becomes this.
$\frac{2+i}{2-i} = \frac{3+4i}{5}$
Now square it. Square the top and the bottom separately.
$\left(\frac{2+i}{2-i}\right)^{2} = \frac{(3+4i)^{2}}{5^{2}}$
Expand the top with $(a+b)^{2}=a^{2}+2ab+b^{2}$ , using $a=3$ and $b=4i$ .
$(3+4i)^{2} = 3^{2}+2(3)(4i)+(4i)^{2}$
Work it out. $(4i)^{2}=16i^{2}=-16$ , so $9+24i-16=-7+24i$ .
$(3+4i)^{2} = 9+24i-16 = -7+24i$
The bottom is $25$ . Put the two together.
$\left(\frac{2-i}{2+i}\right)^{-2} = \frac{-7+24i}{25}$
Split it into a real part and an imaginary part. The real fraction has numerator $-7$ and positive denominator $25$ , so it is negative.
$\left(\frac{2-i}{2+i}\right)^{-2} = -\frac{7}{25}+\frac{24}{25}i$
The real and imaginary parts are
$\operatorname{Re}(z) = -\frac{7}{25}, \qquad \operatorname{Im}(z) = \frac{24}{25}$

Answer

\mathrm{Re}(z)=-\dfrac{7}{25}

and

\mathrm{Im}(z)=\dfrac{24}{25}

Compact solution

Let $z=\left(\frac{2-i}{2+i}\right)^{-2}$ . Take the reciprocal of the nonzero fraction, retaining the positive exponent.

z=\left(\frac{2+i}{2-i}\right)^{2}

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

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

Expand the numerator first.

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

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}

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

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

Now square the simplified quotient. Expand the square, keeping the middle term.

\begin{aligned}z &= (\dfrac{3}{5}+\dfrac{4}{5}i)^2 \\ &= (\dfrac{3}{5})^2+2(\dfrac{3}{5})(\dfrac{4}{5}i)+(\dfrac{4}{5}i)^2 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)^2=a^2+2ab+b^2$}} \\ &= \dfrac{9}{25}+\dfrac{24}{25}i+\dfrac{16}{25}i^2 \\ &= \dfrac{9}{25}+\dfrac{24}{25}i+\dfrac{16}{25}(-1) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^2=-1$}} \\ &= (\dfrac{9}{25}-\dfrac{16}{25})+\dfrac{24}{25}i \\ &= -\dfrac{7}{25}+\dfrac{24}{25}i\end{aligned}

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

\begin{gathered}\operatorname{Re}(z)=-\dfrac{7}{25} \\ \operatorname{Im}(z)=\dfrac{24}{25}\end{gathered}

(vii)

\left(\frac{1-2i}{1+i}\right)^{2}

Solution

Compact solution

Let $z$ denote the given expression. Before squaring, simplify the fraction inside to the form $x+iy$ . Multiply the top and the bottom by the conjugate of the bottom, which is $1-i$ .
$\frac{1-2i}{1+i} = \frac{1-2i}{1+i}\times\frac{1-i}{1-i}$
Work out the numerator, multiplying term by term.
$(1-2i)(1-i) = 1-i-2i+2i^{2}$
Replace $i^{2}$ by $-1$ , so $2i^{2}=-2$ .
$(1-2i)(1-i) = 1-i-2i-2$
Collect. The real numbers give $1-2=-1$ , and the $i$ terms give $-i-2i=-3i$ .
$(1-2i)(1-i) = -1-3i$
Work out the denominator, of the form $(a+b)(a-b)=a^{2}-b^{2}$ .
$(1+i)(1-i) = 1^{2}-i^{2} = 1+1 = 2$
So the fraction inside is this.
$\frac{1-2i}{1+i} = \frac{-1-3i}{2}$
Now square it. Square the top and the bottom separately.
$\left(\frac{1-2i}{1+i}\right)^{2} = \frac{(-1-3i)^{2}}{2^{2}}$
Expand the top with $(a+b)^{2}=a^{2}+2ab+b^{2}$ , using $a=-1$ and $b=-3i$ .
$(-1-3i)^{2} = (-1)^{2}+2(-1)(-3i)+(-3i)^{2}$
Work it out. $(-1)^{2}=1$ , the middle term is $6i$ because minus times minus gives plus, and $(-3i)^{2}=9i^{2}=-9$ .
$(-1-3i)^{2} = 1+6i-9 = -8+6i$
The bottom is $4$ .
$\left(\frac{1-2i}{1+i}\right)^{2} = \frac{-8+6i}{4}$
Divide each term of the top by $4$ and reduce. $\frac{-8}{4}=-2$ and $\frac{6}{4}=\frac{3}{2}$ .
$\left(\frac{1-2i}{1+i}\right)^{2} = -2+\frac{3}{2}i$
The real and imaginary parts are
$\operatorname{Re}(z) = -2, \qquad \operatorname{Im}(z) = \frac{3}{2}$

Answer

\mathrm{Re}(z)=-2

and

\mathrm{Im}(z)=\dfrac{3}{2}

Compact solution

Let $z=\left(\frac{1-2i}{1+i}\right)^{2}$ . Multiply numerator and denominator by the conjugate $1-i$ .

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

Expand the numerator first.

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

Now square the simplified quotient. Expand the square, keeping the middle term.

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

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

\begin{gathered}\operatorname{Re}(z)=-2 \\ \operatorname{Im}(z)=\dfrac{3}{2}\end{gathered}

Common mistake

Watch outWhen you square

\frac{-1-3i}{2}

, square the denominator too:

2^{2}=4

. Using

2

instead of

4

would make the result twice its correct value.

Question 2

Solve the following simultaneous linear equations with complex coefficients for $w$ and $z$ :

(i)

3z+(2+i)w=11-i

and

(2-i)z-w=-1+i

Solution

The given equations are

\begin{aligned}3z+(2+i)w&=11-i,\qquad\cdots(1) \\ (2-i)z-w&=-1+i.\qquad\cdots(2)\end{aligned}

Compact solution

Call the two equations $(1)$ and $(2)$ . In equation $(2)$ the term $-w$ is the simplest to isolate, so make $w$ the subject: add $w$ to both sides and take $-1+i$ across.
$w = (2-i)z-(-1+i)$
A minus in front of a bracket changes the sign of every term inside, so $-(-1+i)=1-i$ . Call this equation $(3)$ .
$w = (2-i)z+1-i \qquad \cdots(3)$
Put this expression for $w$ into equation $(1)$ in place of $w$ .
$3z+(2+i)\bigl[(2-i)z+1-i\bigr] = 11-i$
Multiply out $(2+i)(2-i)$ first. It is of the form $(a+b)(a-b)=a^{2}-b^{2}$ , so it equals $4-i^{2}=5$ .
$(2+i)(2-i) = 5$
Multiply out $(2+i)(1-i)$ term by term.
$(2+i)(1-i) = 2-2i+i-i^{2}$
Replace $i^{2}$ by $-1$ and collect. $-i^{2}=1$ , so $2+1=3$ , and $-2i+i=-i$ .
$(2+i)(1-i) = 3-i$
So equation $(1)$ becomes this.
$3z+5z+(3-i) = 11-i$
Add the two $z$ terms. $3z+5z=8z$ .
$8z+3-i = 11-i$
Take $3-i$ from both sides. On the right, $11-i-3+i=8$ .
$8z = 8$
Divide both sides by $8$ .
$z = 1$
Put $z=1$ into equation $(3)$ to get $w$ .
$w = (2-i)(1)+1-i$
Work out the right side. $(2-i)(1)=2-i$ , then $2-i+1-i=3-2i$ .
$w = 3-2i$
State both answers.
$z = 1, \qquad w = 3-2i$

Answer

z=1

and

w=3-2i

Compact solution

Label the two original equations.

\begin{aligned}3z+(2+i)w&=11-i\qquad\cdots(1) \\ (2-i)z-w&=-1+i\qquad\cdots(2)\end{aligned}

In equation $(2)$ the term $-w$ is the simplest to isolate, so make $w$ the subject.

\begin{aligned}w &= (2-i)z-(-1+i) \\ &= (2-i)z+1-i\qquad\cdots(3) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{distribute the factor }-1$}}\end{aligned}

Put that into equation $(1)$ in place of $w$ .

\begin{aligned}3z+(2+i)\bigl[(2-i)z+1-i\bigr] &= 11-i\end{aligned}

Work out the first of the two products the bracket makes.

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

Then the second.

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

Put both back into equation $(1)$ .

\begin{aligned}3z+5z+(3-i) &= 11-i \\ 8z+3-i &= 11-i \\ 8z &= 8 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{take $3-i$ from both sides}$}} \\ z &= 1 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{divide both sides by $8$}$}}\end{aligned}

Put $z=1$ into equation $(3)$ .

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

Thus the solution is

\begin{aligned}z &= 1, \qquad w = 3-2i\end{aligned}

Check

Check the answer

\begin{aligned}3(1)+(2+i)(3-2i)&=3+(8-i)=11-i\\(2-i)(1)-(3-2i)&=2-i-3+2i=-1+i\end{aligned}

Substitute the stated pair into both original equations. Each left side gives its own right side, so the pair satisfies the entire system.

(ii)

2z+(3+i)w=9-i

and

-iz-iw=-1+i

Solution

The given equations are

\begin{aligned}2z+(3+i)w&=9-i,\qquad\cdots(1) \\ -iz-iw&=-1+i.\qquad\cdots(2)\end{aligned}

Compact solution

Call the two equations $(1)$ and $(2)$ . Both terms in equation $(2)$ carry the factor $-i$ , so divide every term of $(2)$ by $-i$ to make it simpler.
$\frac{-iz}{-i}+\frac{-iw}{-i} = \frac{-1+i}{-i}$
On the left the $-i$ cancels in each term, leaving $z+w$ .
$z+w = \frac{-1+i}{-i}$
Clear the $i$ from the right side by multiplying the top and the bottom by $i$ .
$z+w = \frac{(-1+i)i}{-i\times i}$
The bottom is $-i\times i=-i^{2}=1$ . The top is $-i+i^{2}=-i-1$ .
$z+w = \frac{-1-i}{1} = -1-i$
So equation $(2)$ has become a simple one. Call it $(3)$ .
$z+w = -1-i \qquad \cdots(3)$
Make $z$ the subject of $(3)$ by taking $w$ from both sides. Call it $(4)$ .
$z = -1-i-w \qquad \cdots(4)$
Put this into equation $(1)$ in place of $z$ .
$2(-1-i-w)+(3+i)w = 9-i$
Multiply out the first bracket. $2$ times each term inside.
$-2-2i-2w+(3+i)w = 9-i$
Collect the two $w$ terms. $-2w+(3+i)w=(1+i)w$ .
$-2-2i+(1+i)w = 9-i$
Add $2+2i$ to both sides. On the right, $9-i+2+2i=11+i$ .
$(1+i)w = 11+i$
Divide both sides by $1+i$ , then clear the $i$ by multiplying the top and the bottom by the conjugate $1-i$ .
$w = \frac{11+i}{1+i}\times\frac{1-i}{1-i}$
Work out the numerator term by term.
$(11+i)(1-i) = 11-11i+i-i^{2}$
Replace $i^{2}$ by $-1$ and collect. $-i^{2}=1$ , so $11+1=12$ , and $-11i+i=-10i$ .
$(11+i)(1-i) = 12-10i$
The denominator is $(1+i)(1-i)=1^{2}-i^{2}=2$ .
$(1+i)(1-i) = 2$
So $w$ is this fraction. Divide each term of the top by $2$ .
$w = \frac{12-10i}{2} = 6-5i$
Put $w=6-5i$ into equation $(4)$ to get $z$ .
$z = -1-i-(6-5i)$
A minus in front of the bracket changes both signs. $-1-6=-7$ and $-i+5i=4i$ .
$z = -7+4i$
State both answers.
$z = -7+4i, \qquad w = 6-5i$

Answer

z=-7+4i

and

w=6-5i

Compact solution

Label the two original equations.

\begin{aligned}2z+(3+i)w&=9-i\qquad\cdots(1) \\ -iz-iw&=-1+i\qquad\cdots(2)\end{aligned}

Both terms of equation $(2)$ carry the factor $-i$ , so divide every term by $-i$ .

\begin{aligned}\frac{-iz}{-i}+\frac{-iw}{-i} &= \frac{-1+i}{-i} \\ z+w &= \frac{-1+i}{-i}\end{aligned}

Clear the $i$ from the right side by multiplying top and bottom by $i$ .

\begin{aligned}z+w &= \frac{(-1+i)i}{-i\times i} \\ &= \frac{-i+i^{2}}{-i^{2}} \\ &= \frac{-i-1}{1} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^{2}=-1$}} \\ &= -1-i \qquad\cdots(3)\end{aligned}

Make $z$ the subject of that.

\begin{aligned}z &= -1-i-w \qquad\cdots(4)\end{aligned}

Put it into equation $(1)$ .

\begin{aligned}2(-1-i-w)+(3+i)w &= 9-i \\ -2-2i-2w+(3+i)w &= 9-i \\ -2-2i+(1+i)w &= 9-i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{$-2w+(3+i)w=(1+i)w$}$}} \\ (1+i)w &= 11+i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{add $2+2i$ to both sides}$}}\end{aligned}

Divide by $1+i$ , clearing the $i$ with the conjugate $1-i$ .

\begin{aligned}w &= \frac{11+i}{1+i} \\ &= \frac{11+i}{1+i}\times\frac{1-i}{1-i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{multiply by }\frac{1-i}{1-i}=1$}} \\ &= \frac{11-11i+i-i^{2}}{1^{2}-i^{2}} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^{2}-b^{2}$}} \\ &= \frac{11-10i+1}{1+1} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^{2}=-1$}} \\ &= \frac{12-10i}{2} \\ &= 6-5i\end{aligned}

Put that into equation $(4)$ .

\begin{aligned}z &= -1-i-(6-5i) \\ &= -1-i-6+5i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{distribute the factor }-1$}} \\ &= -7+4i\end{aligned}

Thus the solution is

\begin{aligned}z &= -7+4i, \qquad w = 6-5i\end{aligned}

Check

Check the answer

\begin{aligned}2(-7+4i)+(3+i)(6-5i)&=(-14+8i)+(23-9i)=9-i\\-i(-7+4i)-i(6-5i)&=(4+7i)+(-5-6i)=-1+i\end{aligned}

Substitute the stated pair into both original equations. Each left side gives its own right side, so the pair satisfies the entire system.

(iii)

z-4w=3i

and

2z+3w=11-5i

Solution

The given equations are

\begin{aligned}z-4w&=3i,\qquad\cdots(1) \\ 2z+3w&=11-5i.\qquad\cdots(2)\end{aligned}

Compact solution

Call the two equations $(1)$ and $(2)$ . In equation $(1)$ the term $z$ stands alone with coefficient $1$ , so make it the subject. Call it $(3)$ .
$z = 3i+4w \qquad \cdots(3)$
Put this into equation $(2)$ in place of $z$ .
$2(3i+4w)+3w = 11-5i$
Multiply out the bracket. $2$ times each term inside.
$6i+8w+3w = 11-5i$
Add the two $w$ terms. $8w+3w=11w$ .
$6i+11w = 11-5i$
Take $6i$ from both sides. On the right, $-5i-6i=-11i$ .
$11w = 11-11i$
Divide both sides by $11$ . Divide each term of the right by $11$ .
$w = 1-i$
Put $w=1-i$ into equation $(3)$ .
$z = 3i+4(1-i)$
Multiply out and collect. $4(1-i)=4-4i$ , then $3i+4-4i=4-i$ .
$z = 4-i$
State both answers.
$z = 4-i, \qquad w = 1-i$

Answer

z=4-i

and

w=1-i

Compact solution

Label the two original equations.

\begin{aligned}z-4w&=3i\qquad\cdots(1) \\ 2z+3w&=11-5i\qquad\cdots(2)\end{aligned}

In equation $(1)$ the letter $z$ stands alone, so make it the subject.

\begin{aligned}z &= 3i+4w \qquad\cdots(3)\end{aligned}

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

\begin{aligned}2(3i+4w)+3w &= 11-5i \\ 6i+8w+3w &= 11-5i \\ 6i+11w &= 11-5i \\ 11w &= 11-11i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{take $6i$ from both sides}$}} \\ w &= 1-i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{divide both sides by $11$}$}}\end{aligned}

Put that value into equation $(3)$ .

\begin{aligned}z &= 3i+4(1-i) \\ &= 3i+4-4i \\ &= 4-i\end{aligned}

Thus the solution is

\begin{aligned}z &= 4-i, \qquad w = 1-i\end{aligned}

Check

Check the answer

\begin{aligned}(4-i)-4(1-i)&=4-i-4+4i=3i\\2(4-i)+3(1-i)&=8-2i+3-3i=11-5i\end{aligned}

Substitute the stated pair into both original equations. Each left side gives its own right side, so the pair satisfies the entire system.

(iv)

z+w=3i

and

2z+3w=2

Solution

The given equations are

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

Compact solution

Call the two equations $(1)$ and $(2)$ . Equation $(1)$ has both coefficients equal to $1$ , so make $z$ the subject. Call it $(3)$ .
$z = 3i-w \qquad \cdots(3)$
Put this into equation $(2)$ in place of $z$ .
$2(3i-w)+3w = 2$
Multiply out the bracket. $2$ times each term inside.
$6i-2w+3w = 2$
Collect the $w$ terms. $-2w+3w=w$ .
$6i+w = 2$
Take $6i$ from both sides.
$w = 2-6i$
Put $w=2-6i$ into equation $(3)$ .
$z = 3i-(2-6i)$
A minus in front of the bracket changes both signs, so $-(2-6i)=-2+6i$ . Then $3i+6i=9i$ .
$z = -2+9i$
State both answers.
$z = -2+9i, \qquad w = 2-6i$

Answer

z=-2+9i

and

w=2-6i

Compact solution

Label the two original equations.

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

Equation $(1)$ has both coefficients equal to $1$ , so make $z$ the subject.

\begin{aligned}z &= 3i-w \qquad\cdots(3)\end{aligned}

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

\begin{aligned}2(3i-w)+3w &= 2 \\ 6i-2w+3w &= 2 \\ 6i+w &= 2 \\ w &= 2-6i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{take $6i$ from both sides}$}}\end{aligned}

Put that value into equation $(3)$ .

\begin{aligned}z &= 3i-(2-6i) \\ &= 3i-2+6i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{distribute the factor }-1$}} \\ &= -2+9i\end{aligned}

Thus the solution is

\begin{aligned}z &= -2+9i, \qquad w = 2-6i\end{aligned}

Check

Check the answer

\begin{aligned}(-2+9i)+(2-6i)&=3i\\2(-2+9i)+3(2-6i)&=-4+18i+6-18i=2\end{aligned}

Substitute the stated pair into both original equations. Each left side gives its own right side, so the pair satisfies the entire system.

(v)

2z+(3+i)w=1

and

-z-(1-i)w=2

Solution

The given equations are

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

Compact solution

Call the two equations $(1)$ and $(2)$ . In equation $(2)$ the term $-z$ is the simplest to isolate, so move the $w$ term across.
$-z = 2+(1-i)w$
Multiply both sides by $-1$ to turn $-z$ into $z$ . Every term on the right changes sign. Call it $(3)$ .
$z = -2-(1-i)w \qquad \cdots(3)$
Put this into equation $(1)$ in place of $z$ .
$2\bigl[-2-(1-i)w\bigr]+(3+i)w = 1$
Multiply out the first bracket. $2$ times each term inside.
$-4-2(1-i)w+(3+i)w = 1$
Work out the coefficient of the first $w$ term. $-2(1-i)=-2+2i$ .
$-2(1-i)w = (-2+2i)w$
Add the two $w$ coefficients. $(-2+2i)+(3+i)=1+3i$ .
$-4+(1+3i)w = 1$
Add $4$ to both sides.
$(1+3i)w = 5$
Divide both sides by $1+3i$ , then clear the $i$ by multiplying the top and the bottom by the conjugate $1-3i$ .
$w = \frac{5}{1+3i}\times\frac{1-3i}{1-3i}$
The numerator is $5(1-3i)=5-15i$ .
$5(1-3i) = 5-15i$
The denominator is $(1+3i)(1-3i)=1^{2}-(3i)^{2}=1+9=10$ .
$(1+3i)(1-3i) = 10$
So $w$ is this fraction. Divide each term of the top by $10$ and reduce.
$w = \frac{5-15i}{10} = \frac{1}{2}-\frac{3}{2}i$
Now use equation $(3)$ . First work out $(1-i)w$ , multiplying term by term.
$(1-i)w = \frac{1}{2}-\frac{3}{2}i-\frac{1}{2}i+\frac{3}{2}i^{2}$
Replace $i^{2}$ by $-1$ , so $\frac{3}{2}i^{2}=-\frac{3}{2}$ . Then $\frac{1}{2}-\frac{3}{2}=-1$ and $-\frac{3}{2}i-\frac{1}{2}i=-2i$ .
$(1-i)w = -1-2i$
Now $z=-2-(1-i)w$ , so take $-1-2i$ away from $-2$ . Subtracting a negative adds it, so $-2-(-1-2i)=-2+1+2i$ .
$z = -2+1+2i$
Collect. $-2+1=-1$ .
$z = -1+2i$
State both answers.
$z = -1+2i, \qquad w = \frac{1}{2}-\frac{3}{2}i$

Answer

z=-1+2i

and

w=\dfrac{1}{2}-\dfrac{3}{2}i

Compact solution

Label the two original equations.

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

In equation $(2)$ the term $-z$ is the simplest to isolate.

\begin{aligned}-z &= 2+(1-i)w \\ z &= -2-(1-i)w\qquad\cdots(3) && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{multiply both sides by $-1$}$}}\end{aligned}

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

\begin{aligned}2\bigl[-2-(1-i)w\bigr]+(3+i)w &= 1 \\ -4-2(1-i)w+(3+i)w &= 1 \\ -4+(-2+2i)w+(3+i)w &= 1 \\ -4+(1+3i)w &= 1 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{$(-2+2i)+(3+i)=1+3i$}$}} \\ (1+3i)w &= 5 && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{add $4$ to both sides}$}}\end{aligned}

Divide by $1+3i$ , clearing the $i$ with the conjugate $1-3i$ .

\begin{aligned}w &= \frac{5}{1+3i} \\ &= \frac{5}{1+3i}\times\frac{1-3i}{1-3i} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{multiply by }\frac{1-3i}{1-3i}=1$}} \\ &= \frac{5-15i}{1^{2}-(3i)^{2}} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$(a+b)(a-b)=a^{2}-b^{2}$}} \\ &= \frac{5-15i}{1+9} && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$i^{2}=-1$}} \\ &= \frac{5-15i}{10} \\ &= \frac{1}{2}-\frac{3}{2}i\end{aligned}

Now back to equation $(3)$ , starting with the term $(1-i)w$ .

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

Put that into equation $(3)$ .

\begin{aligned}z &= -2-(-1-2i) \\ &= -2+1+2i && \fcolorbox{#9BB0DE}{transparent}{\small\textcolor{#4F6FAE}{$\text{taking away a negative adds it}$}} \\ &= -1+2i\end{aligned}

Thus the solution is

\begin{aligned}z &= -1+2i, \qquad w = \frac{1}{2}-\frac{3}{2}i\end{aligned}

Check · Common mistake

Check the answer

\begin{aligned}2(-1+2i)+(3+i)\left(\frac12-\frac32i\right)&=(-2+4i)+(3-4i)=1\\-(-1+2i)-(1-i)\left(\frac12-\frac32i\right)&=(1-2i)-(-1-2i)=2\end{aligned}

Substitute the stated pair into both original equations. Each left side gives its own right side, so the pair satisfies the entire system.

Watch outWhen multiplying both sides of equation

(2)

-1

, every term changes sign, including the

2

on the right. It becomes

z=-2-(1-i)w

, not

z=2-(1-i)w

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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

Real and imaginary parts of a power

Powers of a quotient

Real and imaginary parts from the conjugate

Simultaneous linear equations with complex coefficients

Reading the parts after simplification

Why (x+iy)−1(x+iy)^{-1} has the parts it does

The parts of a square and a reciprocal square

Why substitution solves a system with complex coefficients

Recovering the two parts from conjugates

Example 17

Example 18

Example 19

Example 20

Example 21

Question 1

Question 2

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

Why $(x+iy)^{-1}$ has the parts it does

Class 10
Mathematics