### Polar Form and De Moivre's Theorem

#### Definition

##### Polar Form

In addition to the Cartesian form, $z=a+bi$, complex numbers can also be written in trigonometric polar form $z = r(\cos \theta + i\sin \theta)$ where $r$ is the modulus and $\theta$ is the argument of the number, in radians.

You may see $\mathrm{cis} \theta$ used in calculations with complex numbers - this is a shorthand for $\cos \theta + i \sin \theta$. |center

See the page on Modulus and Argument for a full description of the calculation of $r$ and $\theta$.

Recall Euler's formula, which relates exponentials and trigonometric formulae.

$e^{i \theta} = \cos \theta + i \sin \theta$

Using this, we can also write complex numbers in exponential polar form

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

The angle notation is often used in electronics: $z = r\angle \theta$

##### De Moivre's Theorem

De Moivre's theorem states that for any complex number $z = \cos \theta + i \sin \theta$ and integer $n$, $(\cos \theta + i\sin \theta)^n = \cos(n\theta)+i\sin(n\theta)$ This can be stated in exponential polar form as $( e^{i\theta})^n= e^{in\theta}$

De Moivre's theorem is useful when finding a power of a complex number. To apply the theorem to a complex number, first convert it to polar form, then apply the identity given in the theorem: \begin{align} (a+bi)^n &= \bigl(r(\cos \theta + i\sin \theta)\bigr)^n \\ &= r^n \cdot (\cos \theta + i\sin \theta)^n \\ &= r^n\bigl(\cos(n\theta)+i\sin(n\theta)\bigr) \end{align}

#### Worked Examples

###### Example 1

Express the complex number $z=2-3i$ in polar form.

###### Solution

Find the modulus and the argument.

$r = \lvert z \rvert = \sqrt{2^2+(-3)^2} = \sqrt{13}$

$\theta = \arg z = \tan^{-1}\left(\dfrac{-3}{2}\right) = -0.98 \text{ radians (to 2 d.p.)}$

Note: Remember to check that the value for $\theta$ is correct by drawing an Argand diagram. See Modulus and Argument for more detail on finding the argument. In exponential polar form, $z = \sqrt{13} e^{-0.98i}$

Or, in trigonometic polar form, $z = \sqrt{13}\bigl(\cos(-0.98) + i\sin(-0.98)\bigr)$

###### Example 2

Use De Moivre's Theorem to compute $(-1+4i)^{4}$.

###### Solution

First, find the modulus. \begin{align} \lvert -1+4i \rvert & = \sqrt{(-1)^2 + 4^2}\\ &= \sqrt{17} \end{align}

Then find the argument. If we are not careful then we could calculate as follows: \begin{align} \arg(-1+4i) &= \tan^{-1}\left(\frac{4}{-1}\right)\\ &= \tan^{-1}(-4)\\ &= -1.33 \end{align} But the angle $\theta = -1.33$ radians lies in the fourth quadrant and we see that it does not correspond with the point $(-1,4)$, which lies in the second quadrant. We must add $\pi$ to this to correct it. See Modulus and Argument for a full discussion and plenty of examples.

So the actual value for $\theta$ is: $\theta = \pi + \tan^{-1}(-4)=1.82 \text{ radians (to 2 d.p.)}$

Now we can write the complex number in polar form. $(-1+4i) = \sqrt{17}(\cos 1.82 + i \sin 1.82)$

So \begin{align} (-1+4i)^{4} &= \bigl(\sqrt{17}(\cos 1.82+ i\sin 1.82)\bigr)^{4}\\ &= (\sqrt{17})^{4} \times \bigl(\cos 1.82 + i\sin 1.82)\bigr)^{4}\\ &= 17^2 \times \bigl( \cos (4 \times 1.82)+i\sin (4 \times 1.82)\bigr) & \text{ (by De Moivre's theorem)}\\ &=289\bigl(\cos 7.28 + i \sin 7.28\bigr) \text{ in polar form, or}\\ &=156.9 +242.7i \text{ (to 1 d.p.)} \end{align}

#### Workbooks

These workbooks produced by HELM are good revision aids, containing key points for revision and many worked examples. 