International Baccalaureate · DP 1: Analysis and Approaches HL

Complex Numbers: Algebraic Form

⌚ ~202 min ▢ 23 blocks ✓ 98 exercises ➣ Prerequisites : Equations, Quadratic Functions

I Complex Numbers

I.1 The Number $i$ and the Set of Complex Numbers

Discover

In 1572, the Italian engineer Rafael Bombelli studied the equation $$x^3 = 15x + 4.$$

Check that $x = 4$ is a solution.
Cardano's formula, the great discovery of his century, gives a solution of this equation as $$x = \sqrt[3]{2+\sqrt{-121}} + \sqrt[3]{2-\sqrt{-121}}.$$ The symbol $\sqrt{-121}$ (Cardano's historical notation, which we will soon abandon) names an impossible number: no real number has a negative square. Yet the equation does have the perfectly real solution $4$. Bombelli's idea: dare to compute with an “impossible” number $i$ satisfying $i^2 = -1$, writing $\sqrt{-121} = 11i$, and see whether the formula secretly says $4$.
Admitting the usual rules of algebra together with $i^2=-1$, expand $(2+i)^3$. What do you notice? Conclude what value the formula produces when its two cube roots are chosen as $2+i$ and $2-i$, which differ only in the sign in front of $i$, as the formula requires.

The number system in which Bombelli's computation is legitimate is that of the complex numbers.

Definition — The Imaginary Unit $i$

The imaginary unit, denoted by $i$, is defined as a number such that: $$\textcolor{colordef}{i^2 = -1}$$

Why We Write $i^2 \equal -1$ and Never $i \equal \sqrt{-1}$

The temptation is great to write $i = \sqrt{-1}$. The following chain of equalities shows why this chapter does not use this notation: $$1 = \sqrt{1} = \sqrt{(-1)\times(-1)} = \sqrt{-1}\times\sqrt{-1} = i \times i = i^2 = -1.$$ Every step looks legal, yet the conclusion $1 = -1$ is absurd. The faulty step is the third one: the law $\sqrt{a}\times\sqrt{b} = \sqrt{ab}$ holds only for $a, b \ge 0$ — exactly the hypothesis that comes with it for square roots. The symbol $\sqrt{\ }$ keeps its meaning only for non-negative real numbers.

We define $i$ by the property $\textcolor{colordef}{i^2 = -1}$, and we never apply the laws of $\sqrt{\ }$ to negative numbers.
Moreover, $-1$ has two square roots in the complex numbers, $i$ and $-i$, and nothing distinguishes one from the other: a symbol $\sqrt{-1}$ would pretend to choose between two indistinguishable numbers.

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Definition — Complex Number and Algebraic Form

A complex number is a number of the form $z = a + bi$, where $a$ and $b$ are real numbers and $i$ is the imaginary unit. This way of writing $z$ is called its algebraic form. The set of all complex numbers is denoted by $\C$.

Note

$i$ cannot be a real number, since no real number squared is equal to $-1$.
A real number $a$ is a complex number where $b=0$, written as $a = a+0i$. Thus $\R \subset \C$.
A number where $a=0$, such as $bi$, is called a purely imaginary number.

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Definition — Real and Imaginary Parts

For a complex number $z = a + bi$ (with $a,b \in \R$):

$a$ is the real part of $z$, denoted $\Re(z)$.
$b$ is the imaginary part of $z$, denoted $\Im(z)$.

By definition, $\Re(z)$ and $\Im(z)$ are real numbers.

Example

For $z = 2+3i$:

$\Re(2+3i)=2$
$\Im(2+3i)=3$

Definition — Complex Conjugate

The complex conjugate of $z = a + bi$ (with $a,b\in\R$) is $\conjugate{z} = a - bi$. The complex conjugate of $z$ is often denoted as $\conjugate{z}$ or $z^{*}$.

Example

The complex conjugate of $2+3i$ is $\conjugate{2+3i} = 2-3i$.

Skills to practice

Recognising the Real and Imaginary Parts
Classifying Complex Numbers
Identifying Real and Imaginary Parts
Finding the Conjugate of a Complex Number

II Operations with Complex Numbers

II.1 Calculating with Complex Numbers

The arithmetic of complex numbers follows the standard rules of algebra, together with the key relation $i^2=-1$.

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Definition — Algebraic Operations

Let $z = a+bi$ and $w = c+di$, where $a,b,c,d \in \R$.

Addition: $$\begin{aligned}[t] z + w &= (a+bi) + (c+di) \\ &= (a+c)+(b+d)i. \end{aligned}$$
Subtraction: $$\begin{aligned}[t] z - w &= (a+bi) - (c+di) \\ &= (a-c)+(b-d)i. \end{aligned}$$
Multiplication: $$\begin{aligned}[t] z \times w &= (a+bi)(c+di) \\ &= ac + adi + bci + bdi^2 \\ &= ac + adi + bci - bd && \text{($i^2 = -1$)}\\ &= (ac-bd)+ (ad+bc)i. \end{aligned}$$
Division (with $w \neq 0$, so that $c^2+d^2 \neq 0$ since $c$ and $d$ are not both zero): $$\begin{aligned}[t] \dfrac{z}{w} &= \frac{a+bi}{c+di} \\ &= \left(\frac{a+bi}{c+di}\right)\left(\frac{c-di}{c-di}\right) && \text{(multiply by 1)}\\ &= \frac{(a+bi)(c-di)}{(c+di)(c-di)} \\ &= \frac{ac-adi+bci-bdi^2}{c^2-(di)^2} && \text{(expand)}\\ &= \frac{ac-adi+bci+bd}{c^2+d^2} && \text{($i^2 = -1$)}\\ &= \frac{(ac+bd) + (bc-ad)i}{c^2+d^2}. \end{aligned}$$

Example

Write $(2+3i)(4-i)$ in the form $a+bi$.

Answer

$$\begin{aligned}[t] (2+3i)(4-i) &= 8 - 2i + 12i - 3i^2 && \text{(expand)}\\ &= 8 + 10i - 3i^2 && \text{(collect like terms)}\\ &= 8 + 10i - 3 \times (-1) && \text{($i^2 = -1$)}\\ &= 11 + 10i. \end{aligned}$$

Proposition — Inverse of a Nonzero Complex Number and Zero Product

Let $z = a + bi$ with $a, b \in \R$ and $z \neq 0$ (so that $a^2 + b^2 \neq 0$). Then $z$ has an inverse, the complex number $\dfrac{1}{z}$ such that $z \times \dfrac{1}{z} = 1$, and $$\textcolor{colorprop}{\frac{1}{z} = \frac{a - bi}{a^2 + b^2} = \frac{a}{a^2+b^2} - \frac{b}{a^2+b^2}\,i.}$$ Zero product. For all complex numbers $z$ and $w$: $$zw = 0 \;\Leftrightarrow\; z = 0 \text{ or } w = 0.$$

Proof

The inverse. Since $a$ and $b$ are not both zero, $a^2 + b^2 > 0$. Moreover $$\begin{aligned}[t] (a + bi)(a - bi) &= a^2 - (bi)^2 && \text{($(x+y)(x-y) = x^2 - y^2$)}\\ &= a^2 - b^2 i^2 && \text{(power of a product)}\\ &= a^2 + b^2. && \text{($i^2 = -1$)} \end{aligned}$$ Dividing by the nonzero real number $a^2 + b^2$ gives $z \times \dfrac{a - bi}{a^2+b^2} = 1$: the number $\dfrac{a - bi}{a^2+b^2}$ is an inverse of $z$. It is the only one: if $z z' = 1$ and $z z'' = 1$, then $z' = z'(z z'') = (z' z) z'' = z''$.
Zero product. If $z = 0$ or $w = 0$, then $zw = 0$, by the usual rules of algebra. Conversely, suppose $zw = 0$ and $z \neq 0$. Multiplying both sides by $\dfrac{1}{z}$: $$\frac{1}{z}(zw) = \frac{1}{z} \times 0 = 0 \quad \text{and} \quad \frac{1}{z}(zw) = \left(\frac{1}{z} z\right) w = w,$$ so $w = 0$.

Example

Write $\dfrac{1}{2+3i}$ and $\dfrac{1}{2i}$ in algebraic form.

Answer

For $z = 2 + 3i$, $a = 2$ and $b = 3$, so $a^2 + b^2 = 4 + 9 = 13$: $$\begin{aligned}[t] \frac{1}{2+3i} &= \frac{2 - 3i}{13} && \text{(inverse of $a + bi$)}\\ &= \frac{2}{13} - \frac{3}{13}\,i. \end{aligned}$$ For $z = 2i = 0 + 2i$, $a = 0$ and $b = 2$, so $a^2 + b^2 = 4$: $$\begin{aligned}[t] \frac{1}{2i} &= \frac{0 - 2i}{4} && \text{(inverse of $a + bi$)}\\ &= -\frac{1}{2}\,i. \end{aligned}$$ Check: $(2+3i)\left(\frac{2}{13} - \frac{3}{13}i\right) = \frac{4 + 9}{13} + \frac{-6 + 6}{13}\,i = 1$, and $2i \times \left(-\frac{1}{2}i\right) = -i^2 = 1$.

Method — Dividing by a Complex Number

To write a quotient $\dfrac{z}{w}$ (with $w \neq 0$) in algebraic form, multiply the numerator and the denominator by the conjugate of the denominator (for $w = c + di$, the number $c - di$): the denominator becomes the real number $w\conjugate{w} = c^2 + d^2$. This is the same technique used to rationalize a binomial denominator such as $3+\sqrt{5}$, and the same computation as the inverse $\dfrac{1}{w} = \dfrac{c - di}{c^2 + d^2}$, since $\dfrac{z}{w} = z \times \dfrac{1}{w}$.

Example

Write $\dfrac{3+2i}{1-2i}$ in the form $a+bi$.

Answer

$$\begin{aligned}[t] \frac{3+2i}{1-2i} &= \frac{(3+2i)}{(1-2i)} \times \frac{(1+2i)}{(1+2i)} && \text{(multiply by 1)}\\ &= \frac{3 + 6i + 2i + 4i^2}{1^2 - (2i)^2} && \text{(expand; $(x-y)(x+y) = x^2 - y^2$)}\\ &= \frac{-1 + 8i}{1 + 4} && \text{($i^2 = -1$)}\\ &= -\frac{1}{5} + \frac{8}{5}i. \end{aligned}$$

Proposition — Powers of $i$

The powers of $i$ repeat in a cycle of length $4$: $$\textcolor{colorprop}{i^0 = 1, \quad i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \;\dots}$$ For any natural number $n$, the value of $i^n$ depends only on the remainder of the Euclidean division of $n$ by $4$: writing $n = 4q + r$ with $q$ the quotient and $r \in \{0, 1, 2, 3\}$ the remainder, $$i^n = i^{4q+r} = \left(i^4\right)^q \times i^r = 1^q \times i^r = i^r.$$

Example

Compute $i^{2026}$.

Answer

$$\begin{aligned}[t] i^{2026} &= i^{4 \times 506 + 2} && \text{($2026 = 4 \times 506 + 2$)}\\ &= \left(i^4\right)^{506} \times i^2 && \text{($x^{m+n} = x^m x^n$)}\\ &= 1^{506} \times (-1) && \text{($i^4 = 1$, $i^2 = -1$)}\\ &= -1. \end{aligned}$$

Skills to practice

Calculating with Complex Numbers
Evaluating Polynomial Expressions of a Complex Number
Dividing Complex Numbers
Simplifying Expressions to Find Real and Imaginary Parts
Calculating Powers of the Imaginary Unit
Proving Properties of the Powers of $i$
Solving Linear Equations

II.2 Calculating with Technology

Method — Calculating with Complex Numbers with Technology

A graphing calculator, or a scientific calculator with a complex mode, computes sums, products, quotients and powers of complex numbers.

Set the mode to complex numbers written in the form $a + bi$.
Enter the imaginary unit with its own key, not a variable named $i$, and put each complex number between brackets: $(1 + 2i)(3 - i)$, $(4 + 7i) \div (1 - 2i)$, $(3 - 2i)^4$.
Read the result in the form $a + bi$. Report a part exactly only when it is known to be exact: the screen shows an integer, a fraction or a short decimal, and a check by hand confirms it; otherwise give the real part and the imaginary part each to 3 significant figures.
Check what the question asks for: when an exact form or a working is required, the screen gives the value to reach, and the computation by hand, with the conjugate for a quotient, gives the working.

The mock-up computes $(1 + 2i)(3 - i) = 5 + 5i$ and $\dfrac{4 + 7i}{1 - 2i} = -2 + 3i$.

Example

Use technology to write each number in the form $a + bi$, then check the second one by hand.

$(3 - 2i)^4$;
$\dfrac{2 + 5i}{3 - 4i}$.

Answer

In complex mode, the input $(3 - 2i)^4$ displays $-119 - 120i$. Two integers, confirmed by hand with $(3 - 2i)^2 = 5 - 12i$ and $(5 - 12i)^2 = 25 - 120i + 144i^2 = -119 - 120i$: $(3 - 2i)^4 = -119 - 120i$.
The input $(2 + 5i) \div (3 - 4i)$ displays $-0.56 + 0.92i$: two short decimals, which the computation by hand, with the conjugate $3 + 4i$ of the denominator, confirms to be exact: $$\begin{aligned}[t] \frac{2 + 5i}{3 - 4i} &= \frac{(2 + 5i)(3 + 4i)}{(3 - 4i)(3 + 4i)} && \text{(multiply by the conjugate)}\\ &= \frac{6 + 8i + 15i + 20i^2}{3^2 + 4^2} && \text{(expand; $(3 - 4i)(3 + 4i) = 3^2 + 4^2$)}\\ &= \frac{-14 + 23i}{25} && \text{($i^2 = -1$)}\\ &= -0.56 + 0.92i. \end{aligned}$$ So $\dfrac{2 + 5i}{3 - 4i} = -0.56 + 0.92i$ exactly: the working gives the value displayed on the screen.

Skills to practice

Calculating with Complex Numbers with Technology

III Equality of Complex Numbers

III.1 Equal Complex Numbers

Proposition — Equality of Complex Numbers

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal: for $a, b, c, d \in \R$, $$ a+bi=c+di \;\Leftrightarrow\; a=c \text{ and } b=d. $$

Proof

($\Rightarrow$) Suppose $a+bi=c+di$. Then $$(a-c) + (b-d)i = 0.$$ This implies $a-c = i(d-b)$. If $d \neq b$, then $$i = \frac{a-c}{d-b}.$$ Since $a,b,c,d$ are real, the right-hand side is a real number, while $i$ is not real (because $i^2=-1$ has no real solution). This is a contradiction. Therefore, the assumption $d \neq b$ must be false, which means $d=b$. Substituting this back into $a-c = i(d-b)$ gives $a-c = 0$, so $a=c$.
($\Leftarrow$) Suppose $a=c$ and $b=d$. Then $bi=di$. Adding the equalities $a=c$ and $bi=di$ gives $$a+bi = c+di.$$

Skills to practice

Solving Equations by Equating Real and Imaginary Parts

IV Properties of the Complex Conjugate

IV.1 Calculating with Conjugates

Proposition — Properties of the Conjugate

For all complex numbers $z$ and $w$: $$\begin{aligned}[t] \conjugate{\conjugate{z}} &= z, \qquad \conjugate{-z} = -\conjugate{z},\\ \conjugate{z + w} &= \conjugate{z} + \conjugate{w}, \qquad \conjugate{z - w} = \conjugate{z} - \conjugate{w},\\ \conjugate{zw} &= \conjugate{z} \; \conjugate{w},\\ \conjugate{\left(\frac{1}{z}\right)} &= \frac{1}{\conjugate{z}} \quad \text{if } z \neq 0, \qquad \conjugate{\left(\frac{z}{w}\right)} = \frac{\conjugate{z}}{\conjugate{w}} \quad \text{if } w \neq 0,\\ \conjugate{z^{\,n}} &= \left(\conjugate{z}\right)^{n} \quad \text{for every natural number } n. \end{aligned}$$ Since $z^{-n} = \dfrac{1}{z^{n}}$ for $z \neq 0$, the last property extends to every integer $n$ when $z \neq 0$.

Proof

Let $z = a + bi$ and $w = c + di$ with $a, b, c, d \in \R$.
Conjugate of the conjugate. $\conjugate{\conjugate{z}} = \conjugate{a - bi} = a + bi = z$.
Sum. $$\begin{aligned}[t] \conjugate{z + w} &= \conjugate{(a+c)+(b+d)i} && \text{(addition)}\\ &= (a+c)-(b+d)i && \text{(definition of the conjugate)}\\ &= (a-bi)+(c-di) && \text{(regroup)}\\ &= \conjugate{z}+\conjugate{w}. \end{aligned}$$ Product. On one hand, $zw = (ac-bd)+(ad+bc)i$, so $\conjugate{zw} = (ac-bd)-(ad+bc)i$. On the other hand, $$\begin{aligned}[t] \conjugate{z}\;\conjugate{w} &= (a-bi)(c-di) && \\ &= ac - adi - bci + bdi^2 && \text{(expand)}\\ &= (ac - bd) - (ad+bc)i. && \text{($i^2 = -1$)} \end{aligned}$$ The two results agree: $\conjugate{zw} = \conjugate{z}\;\conjugate{w}$. Taking $w = -1$ gives $\conjugate{-z} = \conjugate{-1}\,\conjugate{z} = -\conjugate{z}$.
Inverse. Let $z \neq 0$; then $\conjugate{z} \neq 0$ too (its real and imaginary parts are $a$ and $-b$, not both zero). Taking the conjugate of both sides of $z \times \dfrac{1}{z} = 1$ and using the product property: $$\conjugate{z} \times \conjugate{\left(\frac{1}{z}\right)} = \conjugate{1} = 1,$$ so $\conjugate{\left(\dfrac{1}{z}\right)}$ is the inverse of $\conjugate{z}$: $\conjugate{\left(\dfrac{1}{z}\right)} = \dfrac{1}{\conjugate{z}}$.
Power. For every natural number $n$, let $P(n)$ be the property « $\conjugate{z^{\,n}} = \left(\conjugate{z}\right)^n$ ».

Base case: with the convention $z^0 = 1$ for every complex number $z$, $\conjugate{z^{\,0}} = \conjugate{1} = 1 = \left(\conjugate{z}\right)^0$, so $P(0)$ is true.
Inductive step: suppose $P(n)$ is true for some natural number $n$. Then $$\begin{aligned}[t] \conjugate{z^{\,n+1}} &= \conjugate{z^{\,n} \times z} && \text{($z^{n+1} = z^n \times z$)}\\ &= \conjugate{z^{\,n}} \; \conjugate{z} && \text{(conjugate of a product)}\\ &= \left(\conjugate{z}\right)^{n} \conjugate{z} && \text{(inductive hypothesis)}\\ &= \left(\conjugate{z}\right)^{n+1}, \end{aligned}$$ so $P(n+1)$ is true.
Conclusion: by induction, $P(n)$ is true for every natural number $n$.

Difference and quotient. Since $z - w = z + (-w)$, the sum and the opposite give $\conjugate{z - w} = \conjugate{z} + \conjugate{-w} = \conjugate{z} - \conjugate{w}$; since $\dfrac{z}{w} = z \times \dfrac{1}{w}$ for $w \neq 0$, the product and the inverse give $\conjugate{\left(\dfrac{z}{w}\right)} = \conjugate{z} \times \conjugate{\left(\dfrac{1}{w}\right)} = \conjugate{z} \times \dfrac{1}{\conjugate{w}} = \dfrac{\conjugate{z}}{\conjugate{w}}$.

Proposition — Conjugate, Real and Imaginary Parts

For any complex number $z = a+bi$ (with $a, b \in \R$): $$\begin{aligned}[t] z\conjugate{z} &= a^2 + b^2, \\ z + \conjugate{z} &= 2\,\Re(z), \\ z - \conjugate{z} &= 2i\,\Im(z). \end{aligned}$$ In particular:

$z$ is a real number if and only if $\conjugate{z} = z$;
$z$ is purely imaginary if and only if $\conjugate{z} = -z$.

Proof

Let $z = a+bi$ with $a, b \in \R$. $$\begin{aligned}[t] z\conjugate{z} &= (a+bi)(a-bi) && \\ &= a^2 - (bi)^2 && \text{($(x+y)(x-y) = x^2 - y^2$)}\\ &= a^2 - b^2 i^2 && \text{(power of a product)}\\ &= a^2 + b^2. \end{aligned}$$ $$\begin{aligned}[t] z + \conjugate{z} &= (a+bi) + (a-bi) \\ &= 2a = 2\,\Re(z),\\ z - \conjugate{z} &= (a+bi) - (a-bi) \\ &= 2bi = 2i\,\Im(z). \end{aligned}$$ Then $\conjugate{z} = z$ is equivalent to $z - \conjugate{z} = 0$, that is $2i\,\Im(z) = 0$, that is $\Im(z) = 0$: this means exactly that $z$ is real. Similarly $\conjugate{z} = -z$ is equivalent to $z + \conjugate{z} = 0$, that is $\Re(z) = 0$: this means exactly that $z$ is purely imaginary.

Note

The identity $z\conjugate{z} = a^2+b^2$ is the engine of complex division: multiplying a denominator by its conjugate always produces a real, nonnegative number (zero only for $z=0$). The geometric viewpoint on complex numbers will call this quantity the square of the modulus of $z$.

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Method — Solving an Equation Involving $z$ and $\conjugate{z}$

If only $\conjugate{z}$ appears (for instance $-\conjugate{z} = 5 - i$): isolate $\conjugate{z}$, then take the conjugate of both sides, since $\conjugate{\conjugate{z}} = z$.
If $z$ and $\conjugate{z}$ both appear (for instance $2z + \conjugate{z} = 3 - 2i$): write $z = a + bi$ with $a, b$ real, so that $\conjugate{z} = a - bi$; expand and gather the real and imaginary parts; equate them with those of the right-hand side, which gives a system of two equations in $a$ and $b$.
Check the solution in the equation.

Skills to practice

Recognising Properties of the Conjugate
Proving Properties of the Complex Conjugate
Solving Complex Equations involving the Conjugate

V Quadratic Equations in $\mathbb{C}$

V.1 Quadratic Equations with Complex Roots

Proposition — Solving equations of the form $z^2 \equal k$

Consider the equation $z^2 = k$, where $k$ is a real number.

If $k > 0$, the equation has two distinct real solutions: $$ z_1 = \sqrt{k} \quad \text{and} \quad z_2 = -\sqrt{k} $$
If $k = 0$, the equation has a unique real solution: $$ z_0 = 0 $$
If $k < 0$, the equation has two conjugate purely imaginary solutions: $$ z_1 = i\sqrt{-k} \quad \text{and} \quad z_2 = -i\sqrt{-k} $$

Proof

In each case the equation $z^2 = k$ is turned into a product equal to zero, and a product of complex numbers is zero if and only if one of its factors is zero.

If $k > 0$: $z^2 - k = z^2 - \left(\sqrt{k}\right)^2 = \left(z - \sqrt{k}\right)\left(z + \sqrt{k}\right)$, so $z^2 = k$ if and only if $z = \sqrt{k}$ or $z = -\sqrt{k}$.
If $k = 0$: $z^2 = z \times z = 0$ if and only if $z = 0$.
If $k < 0$: then $-k > 0$ and, since $i^2 = -1$, $k = -(-k) = i^2 (-k) = \left(i\sqrt{-k}\right)^2$; hence $z^2 - k = \left(z - i\sqrt{-k}\right)\left(z + i\sqrt{-k}\right)$, and $z^2 = k$ if and only if $z = i\sqrt{-k}$ or $z = -i\sqrt{-k}$.

Proposition — Solving Quadratic Equations with Real Coefficients in $\mathbb{C}$

Consider the equation $az^2 + bz + c = 0$, where $a, b,$ and $c$ are real numbers and $a \neq 0$.
Let $\Delta = b^2 - 4ac$ be the discriminant of this equation.

If $\Delta > 0$, the equation has two distinct real solutions: $$ z_1 = \frac{-b - \sqrt{\Delta}}{2a} \quad \text{and} \quad z_2 = \frac{-b + \sqrt{\Delta}}{2a} $$
If $\Delta = 0$, the equation has a unique real solution: $$ z_0 = \frac{-b}{2a} $$
If $\Delta < 0$, the equation has two conjugate complex solutions: $$ z_1 = \frac{-b - i\sqrt{-\Delta}}{2a} \quad \text{and} \quad z_2 = \frac{-b + i\sqrt{-\Delta}}{2a} $$
If $\Delta \neq 0$, the quadratic can be factorized as: $az^2 + bz + c = a(z - z_1)(z - z_2)$.

Proof

Let $a, b, c$ be real numbers with $a \neq 0$ and $\Delta = b^2 - 4ac$. For every complex number $z$, the canonical form holds in $\C$ exactly as in $\R$, since it only uses the rules of algebra: $$\begin{aligned}[t] az^2 + bz + c &= a\left[z^2 + \frac{b}{a}z + \frac{c}{a}\right] && \text{(factor out $a \neq 0$)}\\ &= a\left[\left(z + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a}\right] && \text{(complete the square)}\\ &= a\left[\left(z + \frac{b}{2a}\right)^2 - \frac{\Delta}{4a^2}\right]. && \text{($\frac{b^2 - 4ac}{4a^2} = \frac{\Delta}{4a^2}$)} \end{aligned}$$ In each case the bracket factors, and the equation $az^2 + bz + c = 0$ becomes a product equal to zero.

$\Delta > 0$: $\dfrac{\Delta}{4a^2} = \left(\dfrac{\sqrt{\Delta}}{2a}\right)^2$, so, as a difference of two squares, $$az^2 + bz + c = a\left(z + \frac{b}{2a} - \frac{\sqrt{\Delta}}{2a}\right)\left(z + \frac{b}{2a} + \frac{\sqrt{\Delta}}{2a}\right) = a(z - z_2)(z - z_1).$$
$\Delta = 0$: $az^2 + bz + c = a\left(z + \dfrac{b}{2a}\right)^2 = a(z - z_0)^2$, a product of two equal factors.
$\Delta < 0$: $-\Delta > 0$ and $\dfrac{\Delta}{4a^2} = -\dfrac{-\Delta}{4a^2} = i^2\,\dfrac{-\Delta}{4a^2} = \left(\dfrac{i\sqrt{-\Delta}}{2a}\right)^2$, so, again as a difference of two squares, $$az^2 + bz + c = a\left(z + \frac{b}{2a} - \frac{i\sqrt{-\Delta}}{2a}\right)\left(z + \frac{b}{2a} + \frac{i\sqrt{-\Delta}}{2a}\right) = a(z - z_2)(z - z_1),$$ and $z_1$, $z_2$ are conjugates, since they differ only by the sign of their imaginary part.

A product of complex numbers is zero if and only if one of its factors is zero (a nonzero complex number has an inverse). Hence the solutions of $az^2 + bz + c = 0$ are exactly $z_1$ and $z_2$ (or $z_0$ when $\Delta = 0$), and no other complex number; the factorizations obtained along the way are the ones stated.

Example

Solve in $\R$, then in $\C$, the equations $z^2 + 10 = 0$ and $z^2 - 4z + 5 = 0$.

Answer

$z^2 + 10 = 0$ is equivalent to $z^2 = -10$, an equation of the form $z^2 = k$ with $k = -10 < 0$. In $\R$, it has no solution. In $\C$, its solutions are $i\sqrt{10}$ and $-i\sqrt{10}$: $S = \left\{-i\sqrt{10}, i\sqrt{10}\right\}$.
For $z^2 - 4z + 5 = 0$: $a = 1$, $b = -4$, $c = 5$ and $$\Delta = b^2 - 4ac = (-4)^2 - 4 \times 1 \times 5 = 16 - 20 = -4 < 0.$$ In $\R$, there is no solution. In $\C$, there are two conjugate solutions: $$\begin{aligned}[t] z_1 &= \frac{-b - i\sqrt{-\Delta}}{2a} = \frac{4 - i\sqrt{4}}{2} = \frac{4 - 2i}{2} = 2 - i, \\ z_2 &= \frac{-b + i\sqrt{-\Delta}}{2a} = \frac{4 + 2i}{2} = 2 + i, \end{aligned}$$ so $S = \{2 - i, 2 + i\}$ and $z^2 - 4z + 5 = (z - 2 + i)(z - 2 - i)$. Check: $(2 + i)^2 - 4(2 + i) + 5 = 3 + 4i - 8 - 4i + 5 = 0$.

Example

Use technology to solve the equation $2z^2 - 3z + 5 = 0$ in $\C$, then find its solutions exactly.

Answer

The calculator is set to complex mode, then the polynomial solver, degree $2$, receives $a = 2$, $b = -3$ and $c = 5$.

The solutions are $z \approx 0.75 - 1.39i$ and $z \approx 0.75 + 1.39i$. Exactly: $a = 2$, $b = -3$, $c = 5$ are real and $$\Delta = b^2 - 4ac = (-3)^2 - 4 \times 2 \times 5 = 9 - 40 = -31 < 0,$$ so the equation has two conjugate complex solutions: $$z_1 = \frac{-b - i\sqrt{-\Delta}}{2a} = \frac{3 - i\sqrt{31}}{4} \quad \text{and} \quad z_2 = \frac{-b + i\sqrt{-\Delta}}{2a} = \frac{3 + i\sqrt{31}}{4}.$$ Since $\frac{3}{4} = 0.75$ and $\frac{\sqrt{31}}{4} \approx 1.3919411$, the screen displays approximations of these exact values, and its two roots are conjugates, as the resolution in $\C$ predicts.

Skills to practice

Recognising the Solutions of a Quadratic Equation in $\mathbb{C}$
Solving Quadratic Equations with Technology
Solving Quadratic Equations
Factorising Quadratic Polynomials
Solving Equations by Factorisation

VI Going further

Skills to practice

Solving Problems without Technology
Solving Problems with Technology
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