Algebra 2 · N.CN.A.3

Conjugates and dividing complex numbers

Adding, subtracting and multiplying complex numbers always lands back on a + bi. What does it take to divide one by another?

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Adding, subtracting and multiplying complex numbersAlgebra 2

Where you are 5 in Complex numbers

Number9

Complex numbers5

Adding, subtracting and multiplying complex numbers

Conjugates and dividing complex numbers

Imaginary and complex numbers Complex roots come in pairs The complex plane, modulus and argument

When you reach for it

Adding, subtracting and multiplying complex numbers never leaves standard form: the answer is always a + bi with a and b real. A quotient such as

\frac{3 + 4i}{1 - 2i}

is not in that form. It is a perfectly good number, but while i sits below the line you cannot read off its real part or its imaginary part. Nor can you split it as \displaystyle \frac{3}{1} + \frac{4i}{-2i}: a fraction splits over a sum on top, \displaystyle \frac{p + q}{r} = \frac{p}{r} + \frac{q}{r}, never over a sum underneath.

Reach for the conjugate whenever i is in a denominator: dividing two complex numbers, finding a reciprocal, or solving an equation in which the unknown is multiplied by a complex number. The move is the one rationalizing a denominator makes with \displaystyle \frac{1}{3 + \sqrt{2}}: multiply above and below by a partner that turns the bottom into a plain real number. That is no coincidence, since i is itself a square root, \displaystyle \sqrt{-1}.

The procedure

The conjugate

The conjugate of the complex number z = a + bi, with a and b real, is

\overline{z} = a - bi

The real part stays and the imaginary part changes sign. So the conjugate of -6 + 2i is -6 - 2i. Flipping a sign twice undoes it, so \overline{\overline{z}} = z: the two numbers form a conjugate pair, each the conjugate of the other. A real number (b = 0) is its own conjugate, and a pure imaginary number bi has conjugate -bi.

Plot a + bi as the point (a, b), real part across and imaginary part up, as on the complex plane. Flipping the sign of b reflects the point in the real axis.

Figure 1The conjugate is the mirror image across the real axis. Both points are the same distance from 0, and by Pythagoras that distance squared is a^2 + b^2.

Why a number times its conjugate is real

Multiply the pair out, replacing i^2 with -1:

\begin{aligned} (a + bi)(a - bi) &= a^2 - abi + abi - b^2 i^2 \\ &= a^2 - b^2(-1) \\ &= a^2 + b^2 \end{aligned}

The two middle terms are opposites and cancel. This is the difference of squares pattern (p + q)(p - q) = p^2 - q^2 with p = a and q = bi, and its minus sign turns into a plus because q^2 = b^2 i^2 = -b^2. So

z\,\overline{z} = a^2 + b^2

Since a and b are real, three things follow, and the whole method rests on them. The product is real: no i is left. It is never negative, being a sum of two squares. And it is zero only when z = 0, because two squares that are each at least 0 add to 0 only if both are 0. In the figure, a^2 + b^2 is the squared distance from 0 to either point. For example, (-6 + 2i)(-6 - 2i) = 36 + 4 = 40.

Dividing

To divide z by a non-zero w = c + di, multiply top and bottom by \overline{w} = c - di:

\frac{z}{w} = \frac{z}{w}\cdot\frac{\overline{w}}{\overline{w}} = \frac{z\,\overline{w}}{c^2 + d^2}

Each step has a reason.

Multiplying by \displaystyle \frac{\overline{w}}{\overline{w}} changes nothing. That fraction is 1, provided \overline{w} \ne 0, and it is not: w \ne 0, so c and d are not both 0.

The bottom becomes real. w\,\overline{w} = c^2 + d^2, a positive real number, by the fact above.

The top is an ordinary product. z\,\overline{w} multiplies out like any two complex numbers.

Dividing by a real number splits. Once the bottom is a real number r, \displaystyle \frac{p + qi}{r} = \frac{p}{r} + \frac{q}{r}i, which is standard form.

Run it on the quotient from the start. The conjugate of 1 - 2i is 1 + 2i:

\begin{aligned} \frac{3 + 4i}{1 - 2i} &= \frac{(3 + 4i)(1 + 2i)}{(1 - 2i)(1 + 2i)} \\ &= \frac{3 + 6i + 4i + 8i^2}{1^2 + 2^2} \\ &= \frac{-5 + 10i}{5} = -1 + 2i \end{aligned}

Check by multiplying back: (-1 + 2i)(1 - 2i) = -1 + 4i + 4 = 3 + 4i. Dividing z by w asks for the number that, multiplied by w, gives z, so this check is the meaning of the answer, not just a test of it.

Done once with letters, the move gives a formula. It is worth seeing, not memorizing, since the move itself is shorter to carry than the formula:

\frac{a + bi}{c + di} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2}

Reciprocals

Put z = 1. Every non-zero w = c + di has a reciprocal, and it is the conjugate divided by c^2 + d^2:

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

To confirm, multiply by c + di: the top becomes (c + di)(c - di) = c^2 + d^2, which cancels the bottom and leaves 1. As with real numbers, dividing by w is the same as multiplying by \displaystyle \frac{1}{w}.

Worked examples

Example 1

Conjugates and their products

Write the conjugate of each number, then multiply the number by its conjugate.

(a) -1 - 4i \qquad (b) 8i \qquad (c) 7 \qquad (d) \displaystyle \frac{1}{2} - \frac{3}{2}i

Solution

Keep the real part, flip the sign of the imaginary part, and use z\,\overline{z} = a^2 + b^2 with the real numbers a and b.

(a) Here a = -1 and b = -4. The conjugate is -1 + 4i, and the product is (-1)^2 + (-4)^2 = 17.

(b) 8i = 0 + 8i, so the conjugate is -8i. The product is 0^2 + 8^2 = 64; directly, (8i)(-8i) = -64i^2 = 64.

(d) The conjugate is \displaystyle \frac{1}{2} + \frac{3}{2}i, and the product is \displaystyle \frac{1}{4} + \frac{9}{4} = \frac{5}{2}. It is real, though not a whole number; nothing requires that.

Example 2

A real number divided by a complex one

Write \displaystyle \frac{10}{3 + i} in standard form.

Solution

The conjugate of the bottom is 3 - i. Multiply above and below by it:

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

The bottom came to 10, which cancels the 10 on top. Check: (3 - i)(3 + i) = 9 + 1 = 10.

Example 3

An imaginary number on top

Write \displaystyle \frac{5i}{2 - i} in standard form.

Solution

Multiply above and below by 2 + i, the conjugate of the bottom:

\begin{aligned} \frac{5i}{2 - i} &= \frac{5i(2 + i)}{(2 - i)(2 + i)} \\ &= \frac{10i + 5i^2}{2^2 + 1^2} \\ &= \frac{-5 + 10i}{5} = -1 + 2i \end{aligned}

This is the same number as \displaystyle \frac{3 + 4i}{1 - 2i} in the procedure. Two fractions with different tops and bottoms can be equal, and the test is the one for ordinary fractions, cross-multiplying: 5i(1 - 2i) = 10 + 5i and (3 + 4i)(2 - i) = 10 + 5i.

Example 4

Dividing by a pure imaginary number

Write in standard form (a) \displaystyle \frac{1}{i} \qquad (b) \displaystyle \frac{8 + 6i}{2i}

Solution

(a) The conjugate of i = 0 + 1i is -i, and i(-i) = -i^2 = 1:

\frac{1}{i} = \frac{1 \cdot (-i)}{i \cdot (-i)} = \frac{-i}{1} = -i

So the reciprocal of i is -i. Check: i(-i) = -i^2 = 1.

(b) The conjugate -2i would work, but so does plain i: a pure imaginary bottom times i is already real, because i^2 = -1.

\begin{aligned} \frac{8 + 6i}{2i} &= \frac{(8 + 6i)\,i}{2i \cdot i} = \frac{8i + 6i^2}{2i^2} \\ &= \frac{-6 + 8i}{-2} = 3 - 4i \end{aligned}

Check: 2i(3 - 4i) = 6i - 8i^2 = 8 + 6i.

Example 5

A reciprocal, and dividing by multiplying

(a) Find the reciprocal of 4 - 3i. (b) Use it to write \displaystyle \frac{2 - i}{4 - 3i} in standard form.

Solution

(a) In the reciprocal formula c = 4 and d = -3, so the bottom is 4^2 + (-3)^2 = 25 and

\frac{1}{4 - 3i} = \frac{4 + 3i}{25} = \frac{4}{25} + \frac{3}{25}i

Check: (4 - 3i)(4 + 3i) = 16 + 9 = 25, and \displaystyle \frac{25}{25} = 1.

(b) Dividing by a number is multiplying by its reciprocal:

\begin{aligned} \frac{2 - i}{4 - 3i} &= (2 - i)\cdot\frac{4 + 3i}{25} \\ &= \frac{8 + 6i - 4i - 3i^2}{25} \\ &= \frac{11 + 2i}{25} = \frac{11}{25} + \frac{2}{25}i \end{aligned}

This is the conjugate method written in a different order. Fractions in the answer are normal: the bottom is 25 and nothing cancels.

Example 6

Solving an equation

Solve (1 + 2i)z - 4i = 7 for the complex number z.

Solution

Treat it as a linear equation. Add 4i to both sides, then divide both sides by 1 + 2i, which is allowed because 1 + 2i \ne 0:

\begin{aligned} (1 + 2i)z &= 7 + 4i \\ z &= \frac{7 + 4i}{1 + 2i} \\ &= \frac{(7 + 4i)(1 - 2i)}{1^2 + 2^2} \\ &= \frac{15 - 10i}{5} = 3 - 2i \end{aligned}

The top multiplies out as 7 - 14i + 4i - 8i^2 = 15 - 10i. Check in the original equation: (1 + 2i)(3 - 2i) = 3 + 4i + 4 = 7 + 4i, and 7 + 4i - 4i = 7.

Example 7

Choosing a number to make a quotient real

Find the real number k for which \displaystyle \frac{k + 6i}{3 - i} is a real number, and find that real number.

Solution

Put the quotient in standard form with k left as a letter. The conjugate of the bottom is 3 + i, and (3 - i)(3 + i) = 10:

\begin{aligned} \frac{k + 6i}{3 - i} &= \frac{(k + 6i)(3 + i)}{10} \\ &= \frac{3k + ki + 18i + 6i^2}{10} \\ &= \frac{3k - 6}{10} + \frac{k + 18}{10}\,i \end{aligned}

Since k is real, the imaginary part is \displaystyle \frac{k + 18}{10}, and the quotient is real exactly when that is 0, so k = -18. The quotient is then \displaystyle \frac{3(-18) - 6}{10} = -6.

Check: -6(3 - i) = -18 + 6i, which is the top when k = -18. The quotient is real because the top is a real multiple of the bottom.

Boundaries

The bottom must not be 0. \displaystyle \frac{z}{0} has no meaning, here as for real numbers, and the method shows it: 0 = 0 + 0i is its own conjugate and 0^2 + 0^2 = 0. Every other complex number has c^2 + d^2 > 0, and so has a reciprocal.

Standard form first. The conjugate flips the sign of b in a + bi with a and b real, so write the number that way before flipping. i(2 + i) = -1 + 2i, so its conjugate is -1 - 2i; flipping the sign inside the brackets instead gives i(2 - i) = 1 + 2i, a different number. A square root of a negative number becomes a multiple of i first: in \displaystyle \frac{6}{\sqrt{-9}} the bottom is 3i, as in standard form.

a and b must be real. The middle terms cancel whatever a and b are; what makes a^2 + b^2 real and positive is that a and b are real numbers.

The conjugate always works; it is not the only multiplier that can. Any non-zero multiplier that makes the bottom real gives the same answer, such as i for a pure imaginary bottom in Example 4. Multiplying by the bottom itself does not work in general: (c + di)^2 = (c^2 - d^2) + 2cd\,i is real only when c or d is 0.

Finished means standard form. \displaystyle \frac{-5 + 10i}{5} is correct but not finished until it is split as -1 + 2i, with each part a fraction in lowest terms.

Where this goes next

The pairing of a + bi with a - bi comes back in the roots of polynomials with real coefficients, which come in conjugate pairs; a negative discriminant already produces one such pair. Rationalizing a denominator is the same move with a square root in place of i. On the complex plane, z\,\overline{z} = a^2 + b^2 is the square of the modulus, the distance from 0, and in polar form division gets a picture: divide the distances and subtract the angles.

Exercises

Conjugates

Write the conjugate of each number, and multiply the number by its conjugate.

5 - 9i
-3 - i
-12i
\displaystyle 2 + \sqrt{3}\,i

Dividing by a real or a pure imaginary number

Write each in standard form.

\displaystyle \frac{15 - 6i}{3}
\displaystyle \frac{7}{i}
\displaystyle \frac{10 - 4i}{2i}

Dividing

Write each quotient in standard form.

\displaystyle \frac{26}{5 - i}
\displaystyle \frac{6}{1 - i}
\displaystyle \frac{4 + 7i}{2 + i}
\displaystyle \frac{1 + i}{1 - i}
\displaystyle \frac{5 - i}{1 + i}
\displaystyle \frac{-1 + 5i}{3 - i}
\displaystyle \frac{-17 + i}{-1 + 3i}

Reciprocals

Find the reciprocal of 1 + 3i in standard form.
Find the reciprocal of -5 + 12i in standard form.
Is \displaystyle \frac{1}{4} + \frac{1}{5}i the reciprocal of 4 + 5i? Decide by multiplying, then find the reciprocal.

Using division

Solve (3 - i)z = 10 + 10i.
Write \displaystyle \frac{2}{1 + i} + \frac{2}{1 - i} in standard form.
Evaluate \displaystyle \frac{x + 1}{x - 1} at x = 2i.

Reasoning

For z = a + bi, show that z + \overline{z} is real and that z - \overline{z} is purely imaginary or 0.
Multiplying top and bottom by i puts \displaystyle \frac{9}{4i} in standard form in one step. Explain why, and explain why the same multiplier does not make the bottom of \displaystyle \frac{4}{3 + 2i} real.
Explain why 0 is the only complex number with no reciprocal.
Taddeo says that a non-zero complex number divided by its own conjugate always gives a number p + qi with p^2 + q^2 = 1, that is, a point at distance 1 from 0. Test it on z = 2 + i, then prove it.
Kristof says that dividing z by w means multiplying z by \displaystyle \frac{1}{w}. Use the reciprocal formula to show that this gives the same result as multiplying top and bottom of \displaystyle \frac{z}{w} by \overline{w}.

Answers

Conjugate 5 + 9i; product 25 + 81 = 106.
Conjugate -3 + i; product 9 + 1 = 10.
Conjugate 12i; product (-12i)(12i) = -144i^2 = 144.
Conjugate \displaystyle 2 - \sqrt{3}\,i; product 4 + 3 = 7.
5 - 2i. The bottom is already real, so split: \displaystyle \frac{15}{3} - \frac{6}{3}i.
-7i. Multiply top and bottom by -i: \displaystyle \frac{7(-i)}{i(-i)} = \frac{-7i}{1}.
-2 - 5i. Multiply top and bottom by i: \displaystyle \frac{10i - 4i^2}{2i^2} = \frac{4 + 10i}{-2}.
5 + i. \displaystyle \frac{26(5 + i)}{5^2 + 1^2} = \frac{26(5 + i)}{26}.
3 + 3i. \displaystyle \frac{6(1 + i)}{1^2 + 1^2} = \frac{6 + 6i}{2}.
3 + 2i. The top is (4 + 7i)(2 - i) = 15 + 10i and the bottom is 2^2 + 1^2 = 5.
i. The top is (1 + i)(1 + i) = 2i and the bottom is 1^2 + 1^2 = 2.
2 - 3i. The top is (5 - i)(1 - i) = 4 - 6i and the bottom is 2.
\displaystyle -\frac{4}{5} + \frac{7}{5}i. The top is (-1 + 5i)(3 + i) = -8 + 14i and the bottom is 10.
2 + 5i. The top is (-17 + i)(-1 - 3i) = 20 + 50i and the bottom is 10.
\displaystyle \frac{1}{10} - \frac{3}{10}i. The reciprocal is \displaystyle \frac{1 - 3i}{1 + 9}.
\displaystyle -\frac{5}{169} - \frac{12}{169}i. The reciprocal is \displaystyle \frac{-5 - 12i}{25 + 144}.
No. The product \displaystyle (4 + 5i)\left(\frac{1}{4} + \frac{1}{5}i\right) is \displaystyle 1 - 1 + \frac{4}{5}i + \frac{5}{4}i = \frac{41}{20}i, not 1. The reciprocal is \displaystyle \frac{4 - 5i}{41}, that is \displaystyle \frac{4}{41} - \frac{5}{41}i.
z = 2 + 4i. Divide by 3 - i: the top (10 + 10i)(3 + i) is 20 + 40i and the bottom is 10. Check: (3 - i)(2 + 4i) = 10 + 10i.
2. \displaystyle \frac{2}{1 + i} = \frac{2(1 - i)}{2} = 1 - i and likewise \displaystyle \frac{2}{1 - i} = 1 + i; the sum is 2.
\displaystyle \frac{3}{5} - \frac{4}{5}i. At x = 2i the quotient is \displaystyle \frac{1 + 2i}{-1 + 2i}. The top times -1 - 2i is 3 - 4i, and the bottom becomes 1 + 4 = 5.
z + \overline{z} = (a + bi) + (a - bi) = 2a, which is real. z - \overline{z} = 2bi, which is pure imaginary when b \ne 0 and is 0 when b = 0.
The bottom 4i is pure imaginary, so 4i \cdot i = 4i^2 = -4 is real and \displaystyle \frac{9}{4i} = \frac{9i}{-4} = -\frac{9}{4}i. For 3 + 2i, multiplying by i gives 3i + 2i^2 = -2 + 3i: i turns the imaginary part real but turns the real part 3 into 3i, so the bottom still has an imaginary part. The conjugate 3 - 2i clears it: \displaystyle \frac{4(3 - 2i)}{13} = \frac{12}{13} - \frac{8}{13}i.
For w = c + di the reciprocal \displaystyle \frac{c - di}{c^2 + d^2} exists whenever c^2 + d^2 \ne 0, and a sum of squares of two real numbers is 0 only when both are 0, that is when w = 0. And 0 has no reciprocal, since 0 \cdot u = 0 \ne 1 for every u.
True. For 2 + i: \displaystyle \frac{2 + i}{2 - i} = \frac{(2 + i)^2}{5} = \frac{3 + 4i}{5}, and \displaystyle \frac{9}{25} + \frac{16}{25} = 1. In general \displaystyle \frac{z}{\overline{z}} = \frac{z^2}{a^2 + b^2} with z^2 = (a^2 - b^2) + 2abi, and (a^2 - b^2)^2 + (2ab)^2 = (a^2 + b^2)^2, so \displaystyle p^2 + q^2 = \frac{(a^2 + b^2)^2}{(a^2 + b^2)^2} = 1.
The reciprocal formula gives \displaystyle \frac{1}{w} = \frac{\overline{w}}{c^2 + d^2}, so \displaystyle z \cdot \frac{1}{w} = \frac{z\,\overline{w}}{c^2 + d^2}. Since w\,\overline{w} = c^2 + d^2, that is exactly \displaystyle \frac{z\,\overline{w}}{w\,\overline{w}}, the result of multiplying top and bottom by \overline{w}.