Conjugate, division, square roots and polynomial equations – Complex Numbers (Senior 6 Mathematics) | Ganzaa

Complex Numbers

Lessons

Senior 6 · Mathematics

Complex Numbers

5 lessons

Lesson 2 of 5

Conjugate, division, square roots and polynomial equations

25 min

A problem Cardano could not avoid

In his book Ars Magna (1545), the Italian mathematician Gerolamo Cardano asked for two numbers whose sum is 10 and whose product is 40. The numbers must be roots of $x^{2} - 10 x + 40 = 0$ , and the formula gives $5 + \sqrt{-15}$ and $5 - \sqrt{-15}$ . Cardano remarked that such arithmetic is "as subtle as it is useless" (in one English translation of his Latin), yet he checked it: $(5 + \sqrt{-15})(5 - \sqrt{-15}) = 25 - (-15) = 40$ . That check multiplies two complex numbers that differ only in the sign of the imaginary part. This lesson makes that multiplication, and division, into routine tools, and then uses them to solve equations.

Multiplication

Expand the brackets as in algebra and replace $i^{2}$ by $- 1$ :

(a + b i) (c + d i) = a c + a d i + b c i + b d i^{2} = (a c - b d) + (a d + b c) i .

For example, $(2 + 3 i) (1 - 4 i) = 2 - 8 i + 3 i - 12 i^{2} = 2 - 5 i + 12 = 14 - 5 i$ . Multiplication in $\mathbb{C}$ is commutative, associative and distributive over addition, with identity $1$ ; together with addition this makes $(\mathbb{C}, +, \times)$ a field, the structure you met in S4 (Binary Operations).

Definition

Conjugate of $z = a + b i$ : the complex number $\bar{z} = a - bi$ , with the same real part and the opposite imaginary part.

In the Argand diagram, $\bar{z}$ is the reflection of $z$ in the real axis.

Example: $\overline{4 - 7i} = 4 + 7i$ , $\overline{5i} = -5i$ and $\bar{6} = 6$ .

The conjugate $\bar{z}$ is the reflection of $z$ in the real axis. Multiplying by $i$ turns $z$ through $90^\circ$ anticlockwise about O and keeps its modulus.

Important

Properties of the conjugate. For all $z = a + b i$ , $z_{1}$ and $z_{2}$ in $\mathbb{C}$ :

$z + \bar{z} = 2a = 2\operatorname{Re}(z)$ and $z - \bar{z} = 2bi = 2i\operatorname{Im}(z)$ .
$z\bar{z} = |z|^2$ , a real number that is never negative.
$\overline{z_1 + z_2} = \bar{z}_1 + \bar{z}_2$ , $\overline{z_1 z_2} = \bar{z}_1\,\bar{z}_2$ and $\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\bar{z}_1}{\bar{z}_2}$ .
$z$ is real if and only if $\bar{z} = z$ ; $|\bar{z}| = |z|$ .

Proof of 2: $z\bar{z} = (a + bi)(a - bi) = a^2 - abi + abi - b^2 i^2 = a^2 + b^2 = |z|^2$ . This is the fact Cardano used.

Inverse and division

Property 2 turns any denominator into a real number. For $z \ne 0$ , multiply the top and bottom by $\bar{z}$ :

\frac{1}{z} = \frac{\bar{z}}{z\bar{z}} = \frac{\bar{z}}{|z|^2} = \frac{a - bi}{a^2 + b^2}, \qquad \frac{z_1}{z_2} = \frac{z_1\,\bar{z}_2}{|z_2|^2} .

So the inverse of $2 + i$ is $\frac{2 - i}{5} = \frac{2}{5} - \frac{1}{5}i$ . Check: $(2 + i) (2 - i) = 5$ , and $\frac{5}{5} = 1$ ✓.

Worked example 1: a quotient in the form $a + b i$

Write $\frac{3 + 4i}{1 - 2i}$ in the form $a + b i$ .

Solution Multiply the top and bottom by the conjugate of the denominator, $1 + 2 i$ :

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

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

Warning

Multiply by the conjugate of the denominator, top and bottom. Multiplying by the denominator itself, or only the numerator, does not clear $i$ from the bottom. Also, $\frac{1}{a + bi}$ is not $\frac{1}{a} + \frac{1}{bi}$ .

Square roots of a complex number

A square root of $z$ is any $w$ with $w^{2} = z$ . To find the square roots of $a + b i$ , put $w = x + y i$ with $x, y$ real. Then $x^{2} - y^{2} + 2 x y i = a + b i$ , so by equality of complex numbers

x^2 - y^2 = a, \qquad 2xy = b .

Taking moduli of $w^{2} = z$ gives a third, easier equation: $∣ w ∣^{2} = ∣ z ∣$ , that is, $x^2 + y^2 = \sqrt{a^2 + b^2}$ . Adding and subtracting the first and third equations gives $x^{2}$ and $y^{2}$ ; the sign of $2 x y = b$ says whether $x$ and $y$ have the same sign or opposite signs. Every non-zero complex number has exactly two square roots, $w$ and $- w$ .

Worked example 2: square roots of $5 - 12 i$

Solution Let $(x + y i)^{2} = 5 - 12 i$ . Then

x^2 - y^2 = 5, \qquad 2xy = -12,

x^2 + y^2 = \sqrt{25 + 144} = 13 .

Adding the first and third: $2 x^{2} = 18$ , so $x^{2} = 9$ . Subtracting: $2 y^{2} = 8$ , so $y^{2} = 4$ . Since $xy = -6 \lt 0$ , $x$ and $y$ have opposite signs: $x = 3, y = - 2$ or $x = - 3, y = 2$ .

The square roots are $\pm(3 - 2i)$ . Check: $(3 - 2 i)^{2} = 9 - 12 i + 4 i^{2} = 5 - 12 i$ ✓.

The same method gives the square roots of $- 3 + 4 i$ as $\pm(1 + 2i)$ . The diagram shows the geometry: the two roots are opposite points, and their modulus is the square root of the modulus of $z$ .

Both square roots of $- 3 + 4 i$ lie on the circle of radius $\sqrt{5}$ (dashed), because $|w|^2 = |{-3 + 4i}| = 5$ . They are opposite points: $w$ and $- w$ .

Equations in $\mathbb{C}$

Linear equations are solved as in $\mathbb{R}$ , finishing with a division: $(1 + i) z = 3 - i$ gives $z = \frac{3 - i}{1 + i} = \frac{(3 - i)(1 - i)}{2} = \frac{2 - 4i}{2} = 1 - 2i$ .

Quadratic equations $a z^{2} + b z + c = 0$ (with $a \ne 0$ ) are solved by the usual formula,

z = \frac{-b \pm \delta}{2a}, \quad \text{where } \delta \text{ is a square root of } \Delta = b^2 - 4ac .

If the coefficients are real and $\Delta \lt 0$ , then $\delta = i\sqrt{-\Delta}$ and the two roots are conjugates. For $z^{2} + 4 z + 13 = 0$ from the start of Lesson 1: $\Delta = -36$ , $\delta = 6i$ , so $z = \frac{-4 \pm 6i}{2} = -2 \pm 3i$ . If the coefficients are complex, find $\delta$ by the square-root method above. In every case the roots satisfy $z_1 + z_2 = -\frac{b}{a}$ and $z_1 z_2 = \frac{c}{a}$ .

Worked example 3: a quadratic with complex coefficients

Solve $z^{2} - (4 + i) z + (5 + 5 i) = 0$ .

Solution $\Delta = (4 + i)^2 - 4(5 + 5i) = 16 + 8i - 1 - 20 - 20i = -5 - 12i$ .

Square roots of $- 5 - 12 i$ : $x^{2} - y^{2} = - 5$ , $2 x y = - 12$ , $x^{2} + y^{2} = 13$ , so $x^{2} = 4$ , $y^{2} = 9$ and $xy \lt 0$ . Take $\delta = 2 - 3i$ .

z = \frac{(4 + i) \pm (2 - 3i)}{2},

z_1 = \frac{6 - 2i}{2} = 3 - i, \qquad z_2 = \frac{2 + 4i}{2} = 1 + 2i .

Check: $z_{1} + z_{2} = 4 + i$ ✓ and $z_{1} z_{2} = (3 - i) (1 + 2 i) = 3 + 6 i - i + 2 = 5 + 5 i$ ✓. The roots are not conjugates, because the coefficients are not real.

Important

Fundamental theorem of algebra (stated here without proof): every polynomial equation of degree $n \ge 1$ with complex coefficients has exactly $n$ roots in $\mathbb{C}$ , when each root is counted as many times as its multiplicity.

Conjugate root theorem: if $P$ has real coefficients and $P (z_{0}) = 0$ , then $P(\bar{z}_0) = 0$ .

Proof: let $P(z) = a_n z^n + \cdots + a_1 z + a_0$ with every $a_{k}$ real, so $\bar{a}_k = a_k$ . By the properties of the conjugate, $\overline{P(z_0)} = a_n \bar{z}_0^{\,n} + \cdots + a_1 \bar{z}_0 + a_0 = P(\bar{z}_0)$ . Since $P (z_{0}) = 0$ , $P(\bar{z}_0) = \bar{0} = 0$ . ∎

Worked example 4: a cubic with one root given

Given that $1 + 2 i$ is a root of $z^{3} - 5 z^{2} + 11 z - 15 = 0$ , find the other roots.

Solution The coefficients are real, so $1 - 2 i$ is also a root. These two roots give the quadratic factor

(z - 1 - 2 i) (z - 1 + 2 i) = (z - 1)^{2} - (2 i)^{2} = z^{2} - 2 z + 5.

Dividing, $z^{3} - 5 z^{2} + 11 z - 15 = (z^{2} - 2 z + 5) (z - 3)$ . Check by expanding: $z^{3} - 2 z^{2} + 5 z - 3 z^{2} + 6 z - 15 = z^{3} - 5 z^{2} + 11 z - 15$ ✓.

The roots are $1 + 2 i$ , $1 - 2 i$ and $3$ .

A polynomial with real coefficients has its non-real roots in conjugate pairs, placed symmetrically about the real axis.

Tip

After solving any quadratic, check the sum and the product of your roots against $-\frac{b}{a}$ and $\frac{c}{a}$ . It takes thirty seconds and catches most sign errors.

Welch Labs: Imaginary Numbers Are Real, Part 2, A Little History (about 4 min). Watch for the story of Cardano and the cubic equation, and why a formula that gives real answers still needed square roots of negative numbers along the way.

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Practice

Show your working. Questions 7 and 8 are exam length.

Calculate: (a) $(2 + 3 i) (4 - i)$ (b) $(1 - i)^{2}$ (c) $(3 - 2 i) (3 + 2 i)$ .
Write in the form $a + b i$ : (a) $\frac{1}{3 - 4i}$ (b) $\frac{5 + i}{2 - 3i}$ .
Let $z = 2 - i$ . Plot $z$ , $\bar{z}$ and $i z$ on one Argand diagram, and name the transformation that takes $z$ to each of the others.
Find the square roots of (a) $8 + 6 i$ (b) $- 15 - 8 i$ .
Solve $(2 + i) z - 1 = 3 i$ .
Solve $z^{2} - 6 z + 34 = 0$ .
Solve $z^{2} - (3 + 2 i) z + (5 + i) = 0$ , and check your roots with their sum and product.
Given that $2 i$ is a root of $z^{4} + z^{3} + 2 z^{2} + 4 z - 8 = 0$ , solve the equation completely.

Show answersHide answers

(a) $8 - 2 i + 12 i - 3 i^{2} = 11 + 10 i$ (b) $1 - 2 i + i^{2} = - 2 i$ (c) $9 + 4 = 13$ .
(a) $\frac{3 + 4i}{25} = \frac{3}{25} + \frac{4}{25}i$ (b) $\frac{(5 + i)(2 + 3i)}{13} = \frac{7 + 17i}{13} = \frac{7}{13} + \frac{17}{13}i$ .
$\bar{z} = 2 + i$ : reflection in the real axis. $i z = 2 i - i^{2} = 1 + 2 i$ : rotation through $90^\circ$ anticlockwise about O.
(a) $x^{2} - y^{2} = 8$ , $2 x y = 6$ , $x^{2} + y^{2} = 10$ : $x^{2} = 9$ , $y^{2} = 1$ , same signs, so $\pm(3 + i)$ . (b) $x^{2} - y^{2} = - 15$ , $2 x y = - 8$ , $x^{2} + y^{2} = 17$ : $x^{2} = 1$ , $y^{2} = 16$ , opposite signs, so $\pm(1 - 4i)$ .
$z = \frac{1 + 3i}{2 + i} = \frac{(1 + 3i)(2 - i)}{5} = \frac{5 + 5i}{5} = 1 + i$ .
$\Delta = 36 - 136 = -100$ , so $z = \frac{6 \pm 10i}{2} = 3 \pm 5i$ .
$\Delta = (3 + 2i)^2 - 4(5 + i) = -15 + 8i$ , whose square roots are $\pm(1 + 4i)$ . So $z = 2 + 3 i$ or $z = 1 - i$ . Sum $3 + 2 i$ ✓; product $(2 + 3 i) (1 - i) = 5 + i$ ✓.
Real coefficients, so $- 2 i$ is also a root, giving the factor $(z - 2 i) (z + 2 i) = z^{2} + 4$ . Dividing: $z^{4} + z^{3} + 2 z^{2} + 4 z - 8 = (z^{2} + 4) (z^{2} + z - 2) = (z^{2} + 4) (z + 2) (z - 1)$ . The roots are $2 i, - 2 i, 1, - 2$ .

Sources and credits

Written by Ganzaa: the explanations, the diagrams, the worked examples, the practice items and the quiz questions. The historical problem (two numbers with sum 10 and product 40) is from Cardano's Ars Magna (1545), which is in the public domain.
Image: Title page of Ars Magna by Gerolamo Cardano, scan by JCSantos, public domain, via Wikimedia Commons.
Video: Imaginary Numbers Are Real [Part 2: A Little History], Welch Labs, embedded from YouTube.

1A problem Cardano could not avoid

2Multiplication

3Inverse and division

4Square roots of a complex number

5Equations in C \mathbb{C} C

A problem Cardano could not avoid

Multiplication

Inverse and division

Square roots of a complex number

Equations in $\mathbb{C}$