Further complex numbers

Introduction

Recall from FP1 that a complex number $z$ can be expressed in modulus-argument form

$$ z = r\left(\cos\theta +i\sin\theta\right) $$

where $r=|z|$ and $\theta = \arg(z)$.

In FP2 we extend this notion by expressing complex numbers in exponential form, and using the $\sin$ and $\cos$ relations to come up with new trig formulae.

Exponential form

Let $z$ be a complex number. Then $z$ can be expressed in the form

$$ z = r e^{i\theta} $$

where $r=|z|$ and $\theta = \arg(z)$.

Proof (click to expand)

The Maclaurin series for $\cos\theta $, $\sin\theta$, and $e^{x}$ are

$$ \begin{align} \cos\theta &= 1 - \frac{\theta^{2}}{2!} + \frac{\theta^{4}}{4!} - \frac{\theta^{6}}{6!} + \ldots \\ \sin\theta &= \theta - \frac{\theta^{3}}{3!} + \frac{\theta^{5}}{5!} - \frac{\theta^{7}}{7!} + \ldots \\ e^{x} &= 1+ x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\ldots \end{align} $$

Let $x=i\theta$. Then

$$ \begin{align} e^{i\theta} &= 1+ i\theta+\frac{\left(i\theta\right)^{2}}{2!}+\frac{\left(i\theta\right)^{3}}{3!}+\frac{\left(i\theta\right)^{4}}{4!}+\frac{\left(i\theta\right)^{5}}{5!}+\ldots \\ &= 1 + i\theta-\frac{\theta^{2}}{2!}-\frac{i\theta^{3}}{3!}+\frac{\theta^{4}}{4!}+\frac{i\theta^{5}}{5!}+\ldots \\ &= 1 -\frac{\theta^{2}}{2!} +\frac{\theta^{4}}{4!} + \ldots + i\left[\theta -\frac{\theta^{3}}{3!}+\frac{\theta^{5}}{5!}+\ldots \right] \\ &= \cos\theta + i\sin\theta ~ \blacksquare \end{align} $$

This relationship $e^{i\theta} = \cos\theta + i\sin\theta$ is called Euler's formula.

As with the modulus-argument form, the exponential form makes complex number multiplication and division extremely convenient. Take two complex numbers $z_1$ and $z_2$. Then

$$ z_1 \cdot z_2 = r_1 e^{i\theta_1} \cdot r_2 e^{i\theta_2} = r_1 r_2 e ^{i\left(\theta_1 + \theta_2\right)} $$

And

$$ \frac{z_1}{z_2} = \frac{r_1 e^{i\theta_1}}{r_2 e^{i\theta_2}} = \frac{r_1}{r_2} e ^{i\left(\theta_1 - \theta_2\right)} $$

So the operations just follow the normal behaviour of the $e^x$ function.

Example

Q) Find $\frac{3+4i}{5-12i}$.

A) Firstly $|3+4i| = 5$ and $\arg(3+4i) = 0.927^c$ (3.s.f.). Then $|5-12i| = 13$ and $\arg(5-12i) = -1.18^c$ (3.s.f.). So we have

$$ \frac{3+4i}{5-12i} = \frac{5e^{0.927i}}{13e^{-1.18i}} = \frac{5}{13}e^{2.11i} $$

De Moivre's theorem

De Moivre's theorem states that $$ \left(\cos x + i\sin x \right)^{n} = \cos(nx) + i\sin(nx) $$

Proof (click to expand)

We proved earlier that

$$ e^{ix} = \cos x + i\sin x $$

So

$$ \left(\cos x + i\sin x \right)^{n} = \left(e^{ix}\right)^{n} = e^{inx} = \cos(nx) + i\sin(nx) ~ \blacksquare $$

Example

Q) Simplify $\frac{\left(\cos\frac{\pi}{5} + i\sin\frac{\pi}{5}\right)^{3}}{\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)^{2}}$.

A) By de Moivre's theorem

$$ \frac{\left(\cos\frac{\pi}{5} + i\sin\frac{\pi}{5}\right)^{3}}{\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)^{2}} = \frac{e^{i\frac{3\pi}{5}}}{e^{i\frac{4\pi}{3}}} = e^{-i\frac{11\pi}{15}} $$

We can also express this in terms of $\cos$ and $\sin$

$$ \begin{align} e^{-i\frac{11}{15}} &= \cos\left(-\frac{11\pi}{15}\right) + i\sin\left(-\frac{11\pi}{15}\right) \\ &= \cos\left(\frac{11\pi}{15}\right) - i\sin\left(\frac{11\pi}{15}\right) \end{align} $$

Since $\cos$ is an even function and $\sin$ is an odd function. The important part is making sure the subjects of $\cos$ and $\sin$ are the same.

Example

Q) Express $(1+i)^{5}$ in the form $a + bi$ where $a$ and $b$ are real numbers.

A) Firstly the modulus of $1+i$ is $\sqrt{2}$ and the argument of $1+i$ is $\tan^{-1}1 = \frac{\pi}{4}$. Therefore $1+i = \sqrt{2}e^{i\frac{\pi}{4}}$.

By de Moivre's theorem,

$$ (1+i)^{5} = \left(\sqrt{2}e^{i\frac{\pi}{4}}\right)^{5} = \sqrt{32}e^{i\frac{5\pi}{4}} $$

And so $a^{2}+b^{2} = 32$. Also $\pi \lt \frac{5\pi}{4} \lt \frac{3\pi}{2}$ so this new complex number is in the third quadrant. That means both $a$ and $b$ are negative.

$$ \frac{5\pi}{4} = \pi + \tan^{-1}\left(\frac{b}{a}\right) $$

$$ \Rightarrow \frac{b}{a} = \tan\left(\frac{\pi}{4}\right) $$

$$ \Rightarrow a = b $$

The complex number with $a$ and $b$ that satisfy all of these conditions is $-4-4i$ since $(-4)^{2}+(-4)^{2}=32$.

Example

Q) Express $\cos(3\theta)$ in terms of $\cos\theta $ and $\sin(3\theta)$ in terms of $\sin\theta$.

A) By de Moivre's theorem, and letting $c=\cos\theta $ and $s=\sin\theta$ for brevity.

$$ \begin{align} \cos(3\theta) + i\sin(3\theta) &= \left(\cos\theta +i\sin\theta\right)^{3} \\ &= c^{3} + 3c^{2}\left(is\right)+3c\left(is\right)^{2} + \left(is\right)^{3} \\ &= c^{3} + 3ic^{2}s - 3cs^{2} - is^{3} \\ &= c^{3} -3cs^{2} + i\left[3c^{2}s-s^{3}\right] \\ &= c^{3} -3c\left(1-c^{2}\right) + i\left[3\left(1-s^{2}\right)s-s^{3}\right] \\ &= 4c^{3}-3c + i\left[3s-4s^{3} \right] \\ &= 4\cos^{3}\theta - 3\cos\theta + i\left[3\sin\theta-4\sin^{3}\theta \right] \end{align} $$

Comparing real and imaginary parts

$$ \begin{align} \cos(3\theta) &= 4\cos^{3}\theta - 3\cos\theta \\ \sin(3\theta) &= 3\sin\theta-4\sin^{3}\theta \end{align} $$

Additional trig formulae

From the formula $e^{in\theta} = \cos(n\theta) + i\sin(n\theta)$ we can derive the following two formulae for $\sin$ and $\cos$.

$$ \begin{align} \sin(n\theta) &= \frac{e^{in\theta} - e^{-in\theta}}{2i} \\ \cos(n\theta) &= \frac{e^{in\theta} + e^{-in\theta}}{2} \end{align} $$

Proof (click to expand)

Suppose the formulae are true. Then

$$ \begin{align} \cos(n\theta) + i\sin(n\theta) &= \frac{e^{in\theta} + e^{-in\theta}}{2} + \frac{e^{in\theta} - e^{-in\theta}}{2} \\ &= \frac{e^{in\theta} + e^{-in\theta} + e^{in\theta} - e^{-in\theta}}{2} \\ &= \frac{2e^{in\theta}}{2} \\ &= e^{in\theta} ~ \blacksquare \end{align} $$

Example

Q) Show that $\sin^{4}x = \frac{1}{8}\cos(4x) -\frac{1}{4}\cos(2x) + \frac{3}{8}$.

A) From the formula we get

$$ \begin{align} \left(2i\sin x \right)^{4} &= \left(e^{ix}-e^{-ix}\right)^{4} \\ &= \left(e^{ix}\right)^{4}+4\left(e^{ix}\right)^{3}\left(-e^{-ix}\right)+6\left(e^{ix}\right)^{2}\left(-e^{-ix}\right)^{2} + 4\left(e^{ix}\right)\left(-e^{-ix}\right)^{3} +\left(-e^{-ix}\right)^{4} \\ &= e^{4ix} - 4e^{3ix-ix} +6e^{2ix-2ix} -4e^{ix-3ix}+e^{-4ix} \\ &= e^{4ix} -4e^{2ix} +6 -4e^{-2ix}+e^{-4ix} \\ &= e^{4ix}+e^{-4ix} -4\left(e^{2ix}+e^{-2ix}\right) +6 \\ &= 2\cos(4x) -4\cos(2x) + 6 \end{align} $$

Therefore, since $\left(2i\right)^{4}=16$

$$ \sin^{4}x = \frac{1}{8}\cos(4x) -\frac{1}{4}\cos(2x) + \frac{3}{8} $$

$n^{th}$ roots of complex numbers

The exponential form of a complex number allows you to easily find the exponents of a complex number and also its roots, owing to the behaviour of the exponential function.

Every complex number has exactly $n$ distinct $n^{th}$ roots. Note that $f(z) = z^{1/n}$ is a "multi-valued" function - you always get $n$ roots.

(Proof to come later.)

One pretty amazing feature of these $n$ roots is that a complex number's $n^{th}$ roots always line up in a perfect circle on the Argand diagram. You will see this materialise in a second.

If we are to find the $n^{th}$ root of some complex number, then let

$$ z^{n} = r e^{(i\theta+2k\pi)} $$

Where $r = |z^{n}|$, $\theta=\arg\left(z^{n}\right)$, and $k=0, 1, 2, \ldots$

The added "$+2k\pi$" comes from the properties of a complex number's argument. For any complex number, going round an entire circle (i.e. an integer multiple of $2\pi$ radians) from its argument on the Argand diagram you arrive back where you started.

Then

$$ z = \sqrt[n]{r e^{(i\theta+2k\pi)}} = \sqrt[n]{r}e^{\left(i\frac{\theta+2k\pi}{n}\right)} $$

And from de Moivre's theorem

$$ z = \sqrt[n]{r}e^{\left(i\frac{\theta+2k\pi}{n}\right)} = \sqrt[n]{r}\left(\cos\left(\frac{\theta+2k\pi}{n}\right)+i\sin\left(\frac{\theta+2k\pi}{n}\right) \right) $$

Then to find all $n$ of the roots we sub in integer values of $k=0, 1, 2, \ldots$ and stop when we have all $n$ distinct roots.

Example

Q) Find the cube roots of $6+8i$.

A) Let $z^3=6+8i$. Then $|z^3|=10$ and $\arg(z^3)=0.927^c$ (3.s.f.). Then using Euler's formula

$$ z^3 = 10e^{i(0.927 + 2k\pi)} $$

Then taking the cube root of both sides

$$ \begin{align} z &= \sqrt[3]{10e^{i(0.927 + 2k\pi)}} \\ &= \sqrt[3]{10}e^{i\frac{0.927 + 2k\pi}{3}} \end{align} $$

$k=0$:

$$ z = \sqrt[3]{10}e^{i\frac{0.927}{3}} = \sqrt[3]{10}e^{0.309i} $$

$k=1$:

$$ z = \sqrt[3]{10}e^{i\frac{0.927+2\pi}{3}} = \sqrt[3]{10}e^{2.4i} $$

$k=2$:

$$ z = \sqrt[3]{10}e^{i\frac{0.927+4\pi}{3}} = \sqrt[3]{10}e^{4.5i} $$

All three of the roots have the same modulus, so they lie on a circle of radius $\sqrt[3]{10}$ and centered at the origin going around the Argand diagram.

Roots of unity

A specialisation of finding $n^{th}$ roots of complex numbers is finding the $n^{th}$ roots of unity, where 'unity' is 1.

Let $$ z^{n} = 1 = e^{2ik\pi} $$ where $k=0, 1, 2, \ldots$. Then taking the $n^{th}$ root of both sides

$$ z = e^{i\frac{2k\pi}{n}} $$

Then to find all $n$ of the roots we sub in integer values of $k=0, 1, 2, \ldots$ and stop when we have all $n$ distinct roots.

Example

Q) Find the fourth roots of unity.

A) Let

$$ z^{4} = 1 = e^{2ik\pi} $$

Where $k=0, 1, 2, 3$. Then take the fourth roots of both sides

$$ z = e^{i\frac{2k\pi}{4}} = e^{i\frac{k\pi}{2}} $$

$k=0$:

$$ z = e^{i\frac{0\pi}{2}} = 1 $$

$k=1$:

$$ z = e^{i\frac{1\pi}{2}} = e^{i\frac{\pi}{2}} $$

$k=2$:

$$ z = e^{i\frac{2\pi}{2}} = e^{i\pi} = -1 $$

$k=3$:

$$ z = e^{i\frac{3\pi}{2}} $$

All four of the roots have the same modulus, so they lie on a circle of radius 1 and centered at the origin going around the Argand diagram.

Loci on the complex plane

It's possible to express different shapes and transformations on the complex plane.

Circle

This expresses all the points equidistant from some point $z_1$ on the complex plane. It can be thought of as a circle of radius $r$ and centered at $z_1$.

$$ |z-z_1|=r $$

Alternatively these expresses a 'shaded in' disc of points of radius $r$ centered at some point $z_1$ on the complex plane.

$$ |z-z_1| \lt r \textrm{ and } |z-z_1| \le r$$

The inclusive equality version also includes the set of points at the edge of the circle.

Example

Q) Sketch $z$ such that $1\lt |z+3+4i| \lt 4$.

A) This is a 'frisbee' shape. Think of a disc radius 4 centered at $-3-4i$ with a disc of radius 1 centered at the same point cut out.

Perpendicular bisector

This expresses a straight line of all the points equidistant from two points $z_1$ and $z_2$ on the complex plane.

$$ |z-z_1| = |z-z_2| $$

Example

Q) Find the locus of $z$ such that $|z-1| = |z-2+i|$.

A) Let $z=x+yi$

$$ \begin{align} |x+yi-1| &= |x+yi-(2-i)| \\ |(x-1)+yi| &= |(x-2)+i(y+1)| \\ \Rightarrow \sqrt{(x-1)^{2}+y^{2}} &= \sqrt{(x-2)^{2}+(y+1)^{2}} \\ \Rightarrow (x-1)^{2}+y^{2} &= (x-2)^{2}+(y+1)^{2} \\ x^{2}-2x+1+y^{2} &= x^{2}-4x+4+y^{2}+2y+1 \\ -2x &= -4x + 4 + 2y \\ \therefore y &= x -2 \end{align} $$

Half lines

This expresses a 'half line' of points with argument $\theta$. This can be quite hard to visualise. Think of a straight line of points going from $z$ on the line to a point $z_1$, such that every $z$ on the line has the specified argument $\theta$.

$$ \arg(z-z_1) = \theta $$

Example

Q) Find the locus of $z$ such that $\arg(z-1-i) = \frac{\pi}{6}$.

A) From the information given $z_1 = 1+i$ and we need $\arg(z-(1+i)) = \frac{\pi}{6}$.

Firstly $\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$. Now let $z=x+yi$

$$ \arg(x+yi-(1+i)) = \arg((x-1)+i\left(y-1\right)) $$

Therefore

$$ \begin{align} \frac{y-1}{x-1} &= \frac{1}{\sqrt{3}} \\ y - 1 &= (x-1)\frac{1}{\sqrt{3}} \\ \Rightarrow y &= 1 - \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}}x \end{align} $$

And this is a half line so it starts from $1+i$ to ensure that the argument stays as $\frac{\pi}{6}$.

Example

Q) Find the locus of $z$ such that $\frac{\pi}{6} \le \arg(z-i) \le \frac{\pi}{3}$.

Firstly $\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$ and $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$. Now let $z=x+yi$

$$ \arg(x+yi-i) = \arg(x+i(y-1)) $$

For the lower bound of the inequality

$$ \begin{align} \frac{y-1}{x} &= \frac{1}{\sqrt{3}} \\ y - 1 &= \frac{1}{\sqrt{3}}x \\ \Rightarrow y &= 1 + \frac{1}{\sqrt{3}}x \end{align} $$

For the upper bound of the inequality

$$ \begin{align} \frac{y-1}{x} &= \sqrt{3} \\ y - 1 &= \sqrt{3}x \\ \Rightarrow y &= 1 + \sqrt{3}x \end{align} $$

So the 'shaded in' argument area is bounded below by the line $y = 1 + \frac{1}{\sqrt{3}}x$ and bounded above by $y = 1 + \sqrt{3}x$.

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