Complex numbers


In A-level maths you are often asked to solve

$$ax^{2}+bx+c = 0$$

or some variation of that, where a, b, and c are real numbers (we say that $a, b, c \in \mathbb{R}$).

The discriminant, $b^{2}-4ac$, would tell you whether the quadratic equation is soluble. In particular you were told that you needed $b^{2}-4ac \ge 0$ to solve the quadratic.

Why do you need that? It comes from the quadratic formula

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$$

For that to make sense you need $b^{2}-4ac \ge 0$, otherwise the square root term makes no sense. That discriminant condition ensures that the quadratic equation has real roots.

That means there does exist an $x$ which solves $ax^{2}+bx+c = 0$, and $x$ is a real number. It's a number like -2, 5, $\frac{2}{5}$, or $\pi$.

But what if $b^{2}-4ac < 0$? What does that mean? Well for starters you would end up with the square root of a negative number.

This might seem a bit crazy, but bear with me for a second. Suppose that square rooting a negative number made some kind of sense. In particular, let's define $i = \sqrt{-1}$, in other words $i^{2} = -1$.

Then we could do things like

$$ \begin{align} x^{2} + 4 &= 0\\ x^{2} &= -4\\ x &= \pm\sqrt{-4}\\ x &= \pm\sqrt{-1}\sqrt{4}\\ x &= \pm2i \end{align} $$

Here $2i$ and $-2i$ are complex numbers. They have allowed us to solve an equation that doesn't have real roots, and on a broader scale they open up a whole new world of possibilities in mathematics.

Here are some other examples of complex numbers

Strictly speaking, a complex number is a combination of a real part and an imaginary part. The real part is the constant by itself, and the imaginary part is the bit multiplied by the imaginary unit $i$.

A complex number that is just the imaginary part, e.g. $10i$, is called an imaginary number.

Every complex number is in the form $x + yi$, where $x,y \in \mathbb{R}$, and we often use $z$ to denote a complex number.

Every complex number $z$ also has a complex conjugate $\bar{z} = x-yi$. The pair consisting of a complex number and its conjugate is called a conjugate pair. You'll see the significance of this later.

Argand diagrams

What does a complex number look like? We draw out complex numbers on a graph called an argand diagram, or the complex plane.

It's like a standard graph, except we use the $x$-axis for the real part, and the $y$-axis for the imaginary part. Here's one, where $x = a$ and $y = b$ (source)

The real axis shown is just the real number line, and the imaginary axis is an addition to it which allows us to visualise complex numbers on a two-dimensional plane.

Absolute value and argument

The absolute value or modulus of a complex number $z = x+yi$ is the length of the straight line from 0 to $z$ on the argand diagram, denoted $|z|$.

We find it using simple pythagoras

$$|z| = \sqrt{x^{2}+y^{2}}$$.

The argument of a complex number $\arg(z)$ is the angle (in radians) between the positive real axis and the complex number. It's anticlockwise-positive.

This number depends on where the complex number is on the argand diagram. The argand diagram is split into four quadrants, each bordered by the real and imaginary axes


Q) $z = 1 + i$. Find the absolute value of $z$. Then find the argument of $z$ and $\bar{z}$.

A) $|z| = \sqrt{1^{2}+1^{2}} = \sqrt{2}$.

$z$ is in the 1st quadrant, so $\arg(z) = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}$.

$\bar{z} = 1 - i$, so $\bar{z}$ is in the 4th quadrant, therefore $\arg(\bar{z}) = 2\pi - \tan^{-1}\left(\frac{1}{1}\right) = \frac{7\pi}{4}$.

The argument of a complex number is anticlockwise-positive. You can however go clockwise from the positive real axis, but the number will be negative.

This is sometimes done for complex numbers in the 3rd or 4th quadrants, so in this example $\arg(\bar{z})$ is simply $-\frac{\pi}{4}$. We're just going clockwise rather than anticlockwise - we're just going to the same complex number by a shorter path.


Q) $z = -5 - 12i$. Find the argument of $z$.

A) $z$ is in the 3rd quadrant, so $\arg(z) = \pi + \tan^{-1}\left(\frac{12}{5}\right) = 4.32^{c}$ (3.s.f.).

Basic operations

Just like real numbers we can add, subtract, multiply, and divide complex numbers

Addition and subtraction

Add/subtract the real bits and imaginary bits together, respectively

$$(a + bi) \pm (c + di) = (a \pm c) + (b \pm d)i$$

Multiplication and division

Multiply them out like you would an algebraic expression

$$ \begin{align} (a + bi)(c + di) &= ac + adi + bci + bdi^{2} \\ &= ac + (ad+bc)i - bd \\ &= (ac-bd) + (ad+bc)i \end{align} $$

Complex number division is a little trickier and a very common exam question.

To find $\frac{z_{1}}{z_{2}}$, multiply by $\frac{\bar{z}_{2}}{\bar{z}_{2}}$ to eliminate the imaginary component from the denominator

$$ \begin{align} \frac{a + bi}{c + di} &= \frac{a + bi}{c + di} \cdot \frac{c - di}{c - di} \\ &= \frac{ac-adi+bci-bdi^{2}}{c^{2} - (di)^{2}} \\ &= \frac{(ac+bd) + (bc-ad)i}{c^{2}+d^{2}} \\ &= \frac{(ac+bd)}{c^{2}+d^{2}} + \frac{(bc-ad)}{c^{2}+d^{2}}i \end{align} $$


Q) Let $z = 2 - 3i$ and $w = \frac{3}{2}i$. Find $z+w$, $z\cdot w$, and $\frac{z}{w}$.

A) $z + w = 2 + (\frac{3}{2} - 3)i = 2 - \frac{3}{2}i$.

$z\cdot w = (2-3i)\cdot\frac{3}{2}i = 3i - 3\cdot\frac{3}{2}i^{2} = \frac{9}{2}+ 3i$.

$\frac{z}{w} = \frac{2 - 3i}{\frac{3}{2}i} = \frac{2 - 3i}{\frac{3}{2}i} \cdot \frac{-\frac{3}{2}i}{-\frac{3}{2}i} = \frac{-\frac{9}{2}- 3i}{\frac{9}{4}} = -2- \frac{4}{3}i$


Complex numbers also have inverses just like real numbers. To find the inverse of some complex number $z = a+bi$ we use the same method for dividing two complex numbers

$$ \begin{align} \frac{1}{a+bi} &= \frac{1}{a+bi} \cdot \frac{a-bi}{a-bi} \\ &= \frac{a-bi}{a^{2} - (bi)^{2}} \\ &= \frac{a-bi}{a^{2}+b^{2}} \\ &= \frac{a}{a^{2}+b^{2}} - \frac{b}{a^{2}+b^{2}}i \\ &= \frac{1}{|z|^{2}}\bar{z} \end{align} $$

(Note from this the important result $z \cdot \bar{z} = |z|^{2}$.)

This is another very common exam question.


Q) $z = -2 - i$. Find $z^{-1}$.

A) $\frac{1}{-2 - i} = \frac{1}{-2 - i} \cdot \frac{-2 + i}{-2 + i} = \frac{-2+i}{4+1} = -\frac{2}{5}+\frac{1}{5}i$.

Properties of the complex conjugate

The complex conjugate also has some general properties that you need to remember for the exam

Modulus-argument form

We can express complex numbers in terms of their moduli and their arguments. This is sometimes more useful than expressing the complex number in the 'regular' form, because it gives us the information about the modulus and argument straight away.

Expressing a complex number $z = x+yi$ on an argand diagram, and using trigonometry, we can see that $x = |z|\cos(\arg(z))$, and $y= |z|\sin(\arg(z))$

Let $z = x + yi$, where $|z| = r$ and $\arg(z) = \theta$. Then the modulus-argument form of $z$ is

$$ \begin{align} z &= |z|\cos(\arg(z)) + |z|\sin(\arg(z))i \\ &= r\cos\theta + ri\sin \theta \\ &= r\left(\cos\theta + i\sin\theta\right) \end{align} $$


Q) Let $z=-2+4i$. Express $z$ in modulus-argument form.

A) $|z| = \sqrt{2^{2}+4^{2}} = 2\sqrt{5}$ and we can say that $\theta = \pi - \tan^{-1}\left(\frac{4}{2}\right) = 2.03^{c}$ (3.s.f.).

Therefore $z = 2\sqrt{5}\left(\cos 2.03 +i\sin 2.03\right)$.

The modulus-argument form is incredibly helpful when it comes to multiplying and dividing complex numbers, an otherwise dull and repetitive task

$$ \begin{align} z_{1} \cdot z_{2} &= r_{1}\left(\cos\theta_{1}+i\sin\theta_{1}\right) \cdot r_{2}\left(\cos\theta_{2}+i\sin\theta_{2}\right) \\ &= r_{1}r_{2}\left(\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})\right) \end{align} $$

Proof (click to expand)

$$ \begin{align} r_{1}\left(\cos\theta_{1}+i\sin\theta_{1}\right) \cdot r_{2}\left(\cos\theta_{2}+i\sin\theta_{2}\right) &= r_{1}r_{2} \left(\cos\theta_{1}+i\sin\theta_{1}\right)\left(\cos\theta_{2}+i\sin\theta_{2}\right) \\ &= r_{1}r_{2}\left(\cos\theta_{1}\cos\theta_{2}+\cos\theta_{1}i\sin\theta_{2} + i\sin\theta_{1}\cos\theta_{2}+i^{2}\sin\theta_{1}\sin\theta_{2}\right) \\ &= r_{1}r_{2}\left(\cos\theta_{1}\cos\theta_{2}-\sin\theta_{1}\sin\theta_{2} + i\left(\sin\theta_{1}\cos\theta_{2} + \cos\theta_{1}\sin\theta_{2}\right)\right) \\ &= r_{1}r_{2}\left(\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})\right) ~ \blacksquare \end{align} $$


$$ \begin{align} \frac{z_{1}}{z_{2}} &= \frac{ r_{1}\left(\cos\theta_{1}+i\sin\theta_{1}\right) }{ r_{2}\left(\cos\theta_{2}+i\sin\theta_{2}\right) } \\ &= \frac{r_{1}}{r_{2}}\left(\cos(\theta_{1}-\theta_{2})+i\sin(\theta_{1}-\theta_{2})\right) \end{align} $$

Proof (click to expand)

$$ \begin{align} \frac{ r_{1}\left(\cos\theta_{1}+i\sin\theta_{1}\right) }{ r_{2}\left(\cos\theta_{2}+i\sin\theta_{2}\right) } &= \frac{r_{1}}{r_{2}} \cdot \frac{\cos\theta_{1}+i\sin\theta_{1}}{\cos\theta_{2}+i\sin\theta_{2}} \\ &= \frac{r_{1}}{r_{2}} \cdot \frac{\cos\theta_{1}+i\sin\theta_{1}}{\cos\theta_{2}+i\sin\theta_{2}} \cdot \frac{\cos\theta_{2}-i\sin\theta_{2}}{\cos\theta_{2}-i\sin\theta_{2}} \\ &= \frac{r_{1}}{r_{2}} \cdot \frac{\cos\theta_{1}\cos\theta_{2}+i\sin\theta_{1}\cos\theta_{2}-i\sin\theta_{2}\cos\theta_{1}-i^{2}\sin\theta_{1}\sin\theta_{2}}{\cos^{2}\theta_{2}-\left(i\sin\theta_{2}\right)^{2}} \\ &= \frac{r_{1}}{r_{2}} \cdot \frac{\cos\theta_{1}\cos\theta_{2}+\sin\theta_{1}\sin\theta_{2}+ i\left(\sin\theta_{1}\cos\theta_{2}-\cos\theta_{1}\sin\theta_{2}\right)}{\cos^{2}\theta_{2}+\sin^{2}\theta_{2}} \\ &= \frac{r_{1}}{r_{2}}(\cos(\theta_{1}-\theta_{2})+i\sin(\theta_{1}-\theta_{2})) ~ \blacksquare \end{align} $$

It's much faster and much more descriptive of what happens when you multiply and divide complex numbers.

Square root of a complex number

Finding the square root of a complex number is another task we can do with some algebra.

Let $z = w^{2}$, where $z=a+bi$ and $w=c+di$ ($w$ is the square root of $z$). Then

$$ \begin{align} a+bi &= (c+di)^{2} \\ a+bi &= c^{2} + 2cdi + (di)^{2}\\ a+bi &= c^{2} - d^{2} + 2cdi \end{align} $$

Comparing real and imaginary parts, we see that (1) $c^{2} - d^{2} = a$ and (2) $2cd = b$.

Sub in $d=\frac{b}{2c}$ to (1) and we get

$$c^{2} - \frac{b^{2}}{4c^{2}} = a$$

Multiply through by $c^{2}$ and rearrange to get

$$ \begin{align} c^{4} - ac^{2} - \frac{b^{2}}{4} &= 0 \\ \left(c^{2}-\frac{a}{2}\right)^{2} - \frac{a^{2}}{4} - \frac{b^{2}}{4} &= 0 \\ \left(c^{2}-\frac{a}{2}\right)^{2} &= \frac{1}{4}(a^{2}+b^{2}) \\ c^{2} &= \frac{a}{2} \pm \frac{1}{2}\sqrt{a^{2}+b^{2}} \\ \end{align} $$

Then sub each value of $c^{2}$ into (1) above to get corresponding $d$.

As you can see this brute force way of doing it is quite long-winded and inefficient. Fortunately in FP2 you will learn a much faster way to do this.

Introduction to complex variables

As we saw in the introduction, complex numbers can act as variables in quadratics just like any other number. Sometimes a quadratic has no real roots and can only be solved by using complex numbers.

In the particular example we used, $x^{2}+4=0$, we saw that the answer wasn't just one complex root, it was two $-2i$ and $2i$, which are complex conjugates of each other.

This isn't an accident. In fact, in polynomials with a complex root $x+yi$, there is always another root $x-yi$, its complex conjugate. You won't ever have just a single complex number by itself as the root of a polynomial. This is called the complex conjugate root theorem.

In simple terms, this is a consequence of the $\pm$ you see when taking the square root in algebraic expressions, and in the quadratic formula used to solve quadratic equations. I have also written a stricter proof below if you're really curious

Proof (click to expand)

Consider the polynomial of degree $n$ with real coefficients $a_{i}$ $$f(x) = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + \ldots + a_{n}x^{n}$$ Suppose that some complex number $z$ is a root, i.e. $f(z)=0$. Then we need to show that $f(\bar{z})=0$ by necessity. $$f(z)=0 \Rightarrow a_{0} + a_{1}z + a_{2}z^{2} + a_{3}z^{3} + \ldots + a_{n}z^{n} = 0$$ In other words, $$\sum\limits_{r=0}^{n} a_{r}z^{r} = 0$$ Also, $$f(\bar{z}) = \sum\limits_{r=0}^{n} a_{r}\left(\bar{z}\right)^{r}$$ By using the properties of conjugacy, $$\sum\limits_{r=0}^{n} a_{r}\left(\bar{z}\right)^{r} = \sum\limits_{r=0}^{n} a_{r}\overline{z^{r}} = \sum\limits_{r=0}^{n} \overline{a_{r}z^{r}} = \overline{\sum\limits_{r=0}^{n} a_{r}z^{r}}$$ But we know that $f(z)=0$, so $$\overline{\sum\limits_{r=0}^{n} a_{r}z^{r}} = \bar{0} = 0$$ Therefore $f(\bar{z}) = 0$ as required $~ \blacksquare$

The complex conjugate root theorem lets you deduce other roots of a polynomial based on any complex roots you might find. If you're given that $2+i$ is the root of a polynomial, you automatically know that $2-i$ is too. It can save a lot of work.


Q) $1+i$ is the root of a quadratic. Show that the quadratic is $z^{2}-2z+2$.

A) $1+i$ is a root, which means that $1-i$ is also a root. A quadratic has at most two distinct roots so we have found all of the roots. Any quadratic can be factorised into the form $(z-\alpha)(z-\beta)$, where $\alpha$ and $\beta$ are roots of the quadratic

$$ \begin{align} f(z) &= (z-(1+i))(z-(1-i)) \\ &= (z-1-i)(z-1+i) \\ &= z^{2}-z+iz-z+1-i-iz+i-i^{2} \\ &= z^{2}-z-z+1+1+i(z-1-z+1) \\ &= z^{2} - 2z +2 +i(0) \\ &= z^{2} -2z+2 \end{align} $$

Solving cubics and quartics

Cubics are polynomials of degree three, and they have at most three distinct roots. Quartics are degree four polynomials, and have at most four distinct roots. In FP1 you will asked to solve a cubic or quartic equation using your knowledge of complex numbers.

The group of roots you will find for a cubic are either three real roots, or one real root and a complex conjugate pair. For quartics, you can either have four real roots, two real roots and a complex conjugate pair, or two complex conjugative pairs. Since this is FP1, there will always be at least one complex conjugate pair in any quadratic you're asked to solve.

In these questions you will be given one of the real or complex roots (usually the latter) to give you a head start. This is the process you should use for solving cubics

  1. Deduce the second root by finding the complex conjugate of the complex root you're given.
  2. Write the polynomial as $(z-\alpha)(z-\beta)(z-\gamma)$, where $\alpha$, $\beta$, and $\gamma$ are the three roots of the cubic.
  3. Sub in the two complex roots that you have found, leaving you with just one unknown root $\gamma$.
  4. Form a quadratic by multiplying out the two factor brackets where you subbed in your complex conjugate root pair. The cubic will now be in the form $(z^{2}-(\alpha+\beta)z+\alpha\beta)(z-\gamma)$.
  5. This quadratic will have a constant value at the end, $\alpha\beta$. The cubic you were asked to solve will also have a constant value at the end. By necessity, the constant value $\alpha\beta$ in the quadratic you just made multiplied by the $-\gamma$ term in the last bracket will be equal to the constant value at the end of the cubic.

Your head might be spinning a little bit so I'll demonstrate it with an example.


Q) Solve $z^{3}-7z^{2}+19z-13 = 0$, given that $3-2i$ is one of the roots.

A) $3-2i$ is a root of the cubic, therefore its conjugate $3+2i$ is also a root of the cubic. We have two out of the three roots we need.

Express the polynomial in terms of its roots

$$ \begin{align} z^{3}-7z^{2}+19z-13 &= (z-3+2i)(z-3-2i)(z-\gamma) \\ &= (z^{2} -3z-2iz-3z+9+6i+2iz-6i-4i^{2})(z-\gamma) \\ &= (z^{2} -3z-3z+9+4 +i(-2+6+2-6))(z-\gamma) \\ &= (z^{2}-6z+13)(z-\gamma) \end{align} $$

We can then deduce that $-13=13\times -\gamma \Rightarrow \gamma = 1$.

The three roots are $3\pm 2i$ and $1$.

When you're asked to solve quartics, the process is basically the same. You form a product of factors $(z-\alpha)(z-\beta)(z-\gamma)(z-\delta)$, where $\alpha$, $\beta$, $\gamma$, $\delta$ are the four roots of the quartic.

In an exam question they will give you one of the complex roots, so you can deduce the second root by taking the conjugate. Again you multiply out the first two brackets using these two roots, forming a quadratic so the quartic looks like $(z^{2}-(\alpha+\beta)z+\alpha\beta)(z-\gamma)(z-\delta)$.

You divide the original quartic you were asked to solve by $(z^{2}-(\alpha+\beta)z+\alpha\beta)$ using algebraic long division, and the result will be the full version of $(z-\gamma)(z-\delta)$, whose roots are $\gamma$ and $\delta$. You can solve this last quadratic and find the last two roots using simple algebra.

You will then have all four of the roots required.

Additional worked examples

Coming soon.

More study materials