Division of Complex Numbers

A2 · P3 · 11 min

Adding, subtracting and multiplying complex numbers is just algebra with $i^2 = -1$ . Dividing needs one extra idea: multiply the top and bottom of the fraction by the conjugate of the denominator, which makes the denominator real. This note explains why that works, sets out the working examiners expect, and then uses division to solve linear and simultaneous equations with complex coefficients. Division appears in almost every P3 complex-numbers question, often as part (a) before a modulus, argument or locus.

Why the conjugate works

A fraction such as $\dfrac{4 + 3i}{2 - i}$ is a perfectly good complex number, but it is not in the form $x + iy$ , because there is an $i$ in the denominator. The task is to move it.

You have met this problem before. To simplify $\dfrac{1}{2 - \sqrt{3}}$ you multiply top and bottom by $2 + \sqrt{3}$ , because $(2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1$ is rational. Complex division is exactly the same trick, with $i$ playing the role of the surd.

The key fact from the introduction is that a complex number times its conjugate is real:

(c + id)(c - id) = c^2 - i^2d^2 = c^2 + d^2.

So multiplying top and bottom by the conjugate of the denominator leaves the value of the fraction unchanged (you are multiplying by $1$ ) and turns the denominator into the real number $c^2 + d^2$ . Dividing a complex number by a real number is easy: divide each part.

Division

For $w = c + id \ne 0$ :

\frac{z}{w} = \frac{z}{w} \times \frac{w^*}{w^*} = \frac{zw^*}{ww^*} = \frac{zw^*}{c^2 + d^2}.

In particular the reciprocal is

\frac{1}{c + id} = \frac{c - id}{c^2 + d^2}.

Dividing two complex numbers

Write down the conjugate of the denominator.
Multiply numerator and denominator by it.
Expand the numerator fully (four terms) and replace $i^2$ by $-1$ .
Work out the denominator as $c^2 + d^2$ ; it must be a real number.
Divide each part by the denominator and write the answer as $x + iy$ , simplifying fractions.

Before you divide, it is worth estimating whether the answer is sensible: $|z/w| = |z|/|w|$ (see polar form), so dividing a number of size $5$ by one of size $\sqrt{5}$ should give something of size $\sqrt{5}$ .

Useful consequences

Conjugates and quotients

$\left(\dfrac{z}{w}\right)^* = \dfrac{z^*}{w^*}$ : the conjugate of a quotient is the quotient of the conjugates.
$\dfrac{1}{z} + \dfrac{1}{z^*} = \dfrac{z^* + z}{zz^*} = \dfrac{2x}{x^2 + y^2}$ , which is real.
If $zz^* = k$ then $\dfrac{k}{z} = z^*$ . For example, for $z = 3 + 4i$ , $\dfrac{25}{z} = 3 - 4i$ .

The last point is a good shortcut. For $z = 3 + 4i$ , the expression $z + \dfrac{25}{z}$ equals $z + z^* = 6$ without any expansion.

Worked examples

A first division

Express $\dfrac{4 + 3i}{2 - i}$ in the form $x + iy$ , showing all your working.

Solution

The conjugate of the denominator is $2 + i$ .

\frac{4 + 3i}{2 - i} \times \frac{2 + i}{2 + i} = \frac{8 + 4i + 6i + 3i^2}{4 - i^2} = \frac{8 + 10i - 3}{4 + 1} = \frac{5 + 10i}{5} = 1 + 2i.

Check by multiplying back: $(1 + 2i)(2 - i) = 2 - i + 4i - 2i^2 = 4 + 3i$ .

A division with a fractional answer

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

Solution

Multiply by $\dfrac{3 - 4i}{3 - 4i}$ :

\frac{(5 - i)(3 - 4i)}{(3 + 4i)(3 - 4i)} = \frac{15 - 20i - 3i + 4i^2}{9 + 16} = \frac{11 - 23i}{25} = \frac{11}{25} - \frac{23}{25}i.

Leave the answer as exact fractions. $0.44 - 0.92i$ is correct but exact form is expected when the question is exact.

A quotient involving the conjugate

Given $z = 1 + 2i$ , express $\dfrac{z}{z^*}$ in the form $x + iy$ , and show that $\left|\dfrac{z}{z^*}\right| = 1$ .

Solution

\frac{z}{z^*} = \frac{1 + 2i}{1 - 2i} \times \frac{1 + 2i}{1 + 2i} = \frac{1 + 4i + 4i^2}{1 + 4} = \frac{-3 + 4i}{5} = -\frac{3}{5} + \frac{4}{5}i.

Its modulus is $\sqrt{\tfrac{9}{25} + \tfrac{16}{25}} = 1$ . This always happens: $|z^*| = |z|$ , so $\left|\dfrac{z}{z^*}\right| = \dfrac{|z|}{|z|} = 1$ .

A parameter that makes a quotient real

The real number $a$ is such that $\dfrac{a + 3i}{1 + i}$ is real. Find $a$ and the value of the quotient.

Solution

\frac{a + 3i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{a - ai + 3i - 3i^2}{2} = \frac{(a + 3) + (3 - a)i}{2}.

The quotient is real when the imaginary part is zero: $3 - a = 0$ , so $a = 3$ . Its value is then $\dfrac{6}{2} = 3$ .

(Similarly it is purely imaginary when $a + 3 = 0$ , that is $a = -3$ , giving $3i$ .)

Solving equations

Linear equations in $z$

If the equation contains $z$ but not $z^*$ , rearrange it exactly as you would a real linear equation. The final step is a division.

A linear equation with complex coefficients

Solve $(2 + i)z + 3 = 8 - 4i$ , giving $z$ in the form $x + iy$ .

Solution

$(2 + i)z = 5 - 4i$ , so

z = \frac{5 - 4i}{2 + i} \times \frac{2 - i}{2 - i} = \frac{10 - 5i - 8i + 4i^2}{5} = \frac{6 - 13i}{5} = \frac{6}{5} - \frac{13}{5}i.

Simultaneous equations

Solve the simultaneous equations

(1 + i)z + w = 2, \qquad z - iw = -1,

giving $z$ and $w$ in the form $x + iy$ .

Solution

From the second equation, $z = -1 + iw$ . Substitute into the first:

(1 + i)(-1 + iw) + w = 2.

Expand: $-1 + iw - i + i^2w + w = -1 - i + iw - w + w = -1 - i + iw$ . So

-1 - i + iw = 2 \quad\Longrightarrow\quad iw = 3 + i \quad\Longrightarrow\quad w = \frac{3 + i}{i}.

Dividing by $i$ : multiply top and bottom by $-i$ (the conjugate of $i$ ), and $i(-i) = 1$ :

w = (3 + i)(-i) = -3i - i^2 = 1 - 3i.

Then $z = -1 + i(1 - 3i) = -1 + i - 3i^2 = 2 + i$ .

Check in the first equation: $(1 + i)(2 + i) + (1 - 3i) = (2 + i + 2i + i^2) + 1 - 3i = 1 + 3i + 1 - 3i = 2$ .

Tip

Dividing by $i$ is the same as multiplying by $-i$ , because $\dfrac{1}{i} = \dfrac{-i}{-i^2} = -i$ . Remember $\dfrac{1}{i} = -i$ ; it saves time.

Equations containing $z$ and $z^*$

If both $z$ and $z^*$ appear, division alone will not separate them. Substitute $z = x + iy$ and equate real and imaginary parts.

Exam style: an equation with z and its conjugate

Solve the equation $(1 + 2i)w + iw^* = 3 + 5i$ , giving your answer in the form $x + iy$ , where $x$ and $y$ are real.

Solution

Let $w = x + iy$ , so $w^* = x - iy$ .

(1 + 2i)(x + iy) = x + iy + 2ix + 2i^2y = (x - 2y) + i(2x + y),

i(x - iy) = ix - i^2y = y + ix.

Adding: $(x - 2y + y) + i(2x + y + x) = (x - y) + i(3x + y)$ .

Equate with $3 + 5i$ :

x - y = 3, \qquad 3x + y = 5.

Adding, $4x = 8$ , so $x = 2$ and $y = -1$ . Hence $w = 2 - i$ .

Exam-hard: an equation with z in a fraction

Find the complex number $z$ satisfying $\dfrac{z + 2i}{z - 3} = 2 - i$ . Give your answer in the form $x + iy$ .

Solution

Multiply both sides by $z - 3$ (with $z \ne 3$ ):

z + 2i = (2 - i)(z - 3) = (2 - i)z - 6 + 3i.

Collect the $z$ terms on one side:

z - (2 - i)z = -6 + 3i - 2i \quad\Longrightarrow\quad (-1 + i)z = -6 + i.

Divide, multiplying by the conjugate $-1 - i$ :

z = \frac{-6 + i}{-1 + i} \times \frac{-1 - i}{-1 - i} = \frac{6 + 6i - i - i^2}{1 + 1} = \frac{7 + 5i}{2} = \frac{7}{2} + \frac{5}{2}i.

Check: $z + 2i = \tfrac{7}{2} + \tfrac{9}{2}i$ and $z - 3 = \tfrac{1}{2} + \tfrac{5}{2}i$ ; then $(2 - i)\left(\tfrac{1}{2} + \tfrac{5}{2}i\right) = 1 + 5i - \tfrac{1}{2}i - \tfrac{5}{2}i^2 = \tfrac{7}{2} + \tfrac{9}{2}i$ .

Common mistakes

Using the conjugate of the numerator

Multiply top and bottom by the conjugate of the denominator. Multiplying by the conjugate of the numerator makes the numerator real and leaves $i$ in the denominator, which achieves nothing.

Getting the denominator wrong

$(3 + 4i)(3 - 4i) = 9 + 16 = 25$ , not $9 - 16 = -7$ . The $-i^2d^2$ term becomes $+d^2$ . The denominator is always $c^2 + d^2$ , a positive real number; if yours is negative or contains $i$ , stop and recheck.

Dividing only part of the numerator

$\dfrac{6 - 13i}{5} = \dfrac{6}{5} - \dfrac{13}{5}i$ , not $\dfrac{6}{5} - 13i$ . Both parts are divided by the denominator.

Trying to divide when z* is present

In $(1 + 2i)w + iw^* = 3 + 5i$ you cannot factor out $w$ , because $w^*$ is a different number. Use $w = x + iy$ and equate parts.

Exam technique

Exam tip

The syllabus states that for division "full details of the working should be shown". The examiner wants to see: the multiplication by $\dfrac{w^*}{w^*}$ , the expanded numerator with $i^2$ visible, and the real denominator. A calculator answer with no working can score zero.
"Express in the form $x + iy$ " means separate the real and imaginary parts at the end: write $\dfrac{11}{25} - \dfrac{23}{25}i$ , not $\dfrac{11 - 23i}{25}$ (though both are usually accepted, the first matches the form exactly).
Check a division by multiplying the answer by the denominator. It takes ten seconds and catches sign errors.
Questions frequently say "find $\dfrac{u}{v}$ in the form $x + iy$ . Hence find the modulus and argument of $\dfrac{u}{v}$ ." The "hence" means use your $x + iy$ ; see the Argand diagram for modulus and argument.

Summary

To divide, multiply top and bottom by the conjugate of the denominator: $\dfrac{z}{w} = \dfrac{zw^*}{ww^*}$ .
$(c + id)(c - id) = c^2 + d^2$ is real and positive, so the denominator becomes real.
Reciprocal: $\dfrac{1}{c + id} = \dfrac{c - id}{c^2 + d^2}$ ; in particular $\dfrac{1}{i} = -i$ .
Show the full working: conjugate, expanded numerator, real denominator, final $x + iy$ .
Linear and simultaneous equations in $z$ : rearrange as usual, then divide.
Equations with both $z$ and $z^*$ : substitute $z = x + iy$ and equate real and imaginary parts.
A quotient is real when its imaginary part is zero and purely imaginary when its real part is zero.

Practice

Question

Express $\dfrac{7 - i}{1 + 2i}$ in the form $x + iy$ .
Express $\dfrac{(1 + i)^2}{3 - i}$ in the form $x + iy$ .
Find $\dfrac{1}{2 - 5i}$ in the form $x + iy$ .
Solve $(1 - 2i)z + 1 = 5 + 3i$ .
Given $z = 3 + 4i$ , show that $z + \dfrac{25}{z}$ is real and find its value.
The real number $k$ is such that $\dfrac{k + i}{2 - i}$ is purely imaginary. Find $k$ .
Solve the simultaneous equations $2z + w = 5 + i$ and $z - iw = -2 - i$ .
Solve $(2 - i)z + 3iz^* = 6 + 4i$ .
Express $\dfrac{1 + i}{1 - i}$ in the form $x + iy$ and hence find $\left(\dfrac{1 + i}{1 - i}\right)^{2027}$ .
The complex number $z = x + iy$ satisfies $\dfrac{1}{z} + \dfrac{1}{z^*} = 1$ and $\operatorname{Im} z = 1$ . Show that $\dfrac{1}{z} + \dfrac{1}{z^*} = \dfrac{2x}{x^2 + y^2}$ and hence find $z$ .

Answers

$\dfrac{(7 - i)(1 - 2i)}{(1 + 2i)(1 - 2i)} = \dfrac{7 - 14i - i + 2i^2}{5} = \dfrac{5 - 15i}{5} = 1 - 3i$ .
$(1 + i)^2 = 2i$ . Then $\dfrac{2i(3 + i)}{(3 - i)(3 + i)} = \dfrac{6i + 2i^2}{10} = \dfrac{-2 + 6i}{10} = -\dfrac{1}{5} + \dfrac{3}{5}i$ .
$\dfrac{2 + 5i}{(2 - 5i)(2 + 5i)} = \dfrac{2 + 5i}{29} = \dfrac{2}{29} + \dfrac{5}{29}i$ .
$(1 - 2i)z = 4 + 3i$ , so $z = \dfrac{(4 + 3i)(1 + 2i)}{5} = \dfrac{4 + 8i + 3i + 6i^2}{5} = \dfrac{-2 + 11i}{5} = -\dfrac{2}{5} + \dfrac{11}{5}i$ .
$\dfrac{25}{3 + 4i} = \dfrac{25(3 - 4i)}{25} = 3 - 4i$ , so $z + \dfrac{25}{z} = (3 + 4i) + (3 - 4i) = 6$ , which is real.
$\dfrac{(k + i)(2 + i)}{5} = \dfrac{2k + ki + 2i + i^2}{5} = \dfrac{(2k - 1) + (k + 2)i}{5}$ . Purely imaginary means real part zero: $2k - 1 = 0$ , $k = \tfrac{1}{2}$ (the imaginary part, $\tfrac{1}{2}$ , is non-zero).
From the first, $w = 5 + i - 2z$ . Substitute: $z - i(5 + i - 2z) = -2 - i$ , so $z - 5i + 1 + 2iz = -2 - i$ , giving $(1 + 2i)z = -3 + 4i$ . Then $z = \dfrac{(-3 + 4i)(1 - 2i)}{5} = \dfrac{-3 + 6i + 4i - 8i^2}{5} = \dfrac{5 + 10i}{5} = 1 + 2i$ and $w = 5 + i - 2 - 4i = 3 - 3i$ .
Let $z = x + iy$ . $(2 - i)(x + iy) = (2x + y) + i(2y - x)$ and $3i(x - iy) = 3y + 3ix$ . Sum: $(2x + 4y) + i(2x + 2y)$ . Real: $2x + 4y = 6$ ; imaginary: $2x + 2y = 4$ . Subtracting, $2y = 2$ , so $y = 1$ , $x = 1$ : $z = 1 + i$ .
$\dfrac{(1 + i)(1 + i)}{(1 - i)(1 + i)} = \dfrac{2i}{2} = i$ . Then $i^{2027}$ : $2027 = 4 \times 506 + 3$ , so the answer is $i^3 = -i$ .
$\dfrac{1}{z} + \dfrac{1}{z^*} = \dfrac{z^* + z}{zz^*} = \dfrac{2x}{x^2 + y^2}$ . With $y = 1$ : $\dfrac{2x}{x^2 + 1} = 1$ , so $x^2 - 2x + 1 = 0$ , $(x - 1)^2 = 0$ , $x = 1$ . Hence $z = 1 + i$ .

Why the conjugate works

Useful consequences

Worked examples

Solving equations

Linear equations in zzz

Equations containing zzz and z∗z^*z∗

Common mistakes

Exam technique

Summary

Practice

Console

Linear equations in $z$

Equations containing $z$ and $z^*$