## Trigonometric form of complex numbers

Let $$z = a + b i$$ denote a complex number with real part $$a$$ and imaginary part $$bi$$.

Then the absolute value or modulus of $$z$$, denoted $$| z |$$, is the distance from $$z$$ to zero in the complex plane and is computed by the formula $$|z|=\sqrt{a^2+b^2}$$.

The line containing $$0$$ and $$z$$ makes an angle $$t$$ with the real axis satisfying $$\tan t=\frac{b}{a}$$. The angle $$t$$ is called the argument of $$z$$ and may be styled $$t=\arg z$$.

Furthermore,

$$\cos t = \dfrac{a}{| z |}$$ and $$\sin t = \dfrac{b}{| z |}$$.

Thus $$a = | z | \cos t$$ and $$b = | z | \sin t$$. So $$z = | z | \cos t + \left(| z | \sin t\right) i$$.

So $$z = | z | (\cos t + i \sin t)$$.

This is the trigonometric form of a complex number.

## Complex product in trigonometric form

Suppose $$z_1$$ and $$z_2$$ are complex numbers. Then their product is

$$\begin{eqnarray*} z_1 \cdot z_2 &=& | z_1 | ( \cos t_1 + i \sin t_1 )\cdot | z_2 |( \cos t_2 + i \sin t_2 )\\ &=& | z_1 | | z_2 | ( \cos t_1 \cos t_2 - \sin t_1 \sin t_2 ) + i ( \cos t_1 \sin t_2 + \sin t_1 \cos t_2)\\ &=& | z_1 | | z_2 | \left(\cos ( t_1 + t_2 ) + i\, \sin ( t_1 + t_2 )\right) \end{eqnarray*}$$

Thus,

$z_1 \cdot z_2 = | z_1 | | z_2 | \left[ \cos ( t_1 + t_2 ) + i\, \sin ( t_1 + t_2 ) \right]$

## Exercise 6.3.1

Let $$z = 3 + 3 i$$.

Find $$|z|$$, the modulus of $$z$$.

Find an angle $$t=\arg z$$ in radians satifying $$z = | z | ( \cos t + i \sin t )$$.

Write $$z$$ in trigonometric form.

See Solution

## Exercise 6.3.2

Show that $$z^2 = | z |^2 ( \cos 2t + i\,\sin 2t )$$.

See Solution

## DeMoivre's Theorem

The result in Exercise 6.3.2 can be extended to the general result, called DeMoivre's Theorem:

If $$z = | z | ( \cos t + i\, \sin t )$$, then $$z^n = | z |^n [ \cos ( n t ) + i\, \sin ( n t ) ]$$.

This means that the modulus of $$z^n$$ is $$|z|^n$$ and $$\arg z^n=n\arg z$$.

Example: Find $$( -1 + 2 i )^5$$

Let $$z = -1 + 2 i$$. Then $$\mod z^5=|z|^5=\left(\sqrt{5}\right)^5\approx55.9$$.

Next, we must find the angle $$t=\arg(-1+2 i)$$.

Since $$z$$ lies in the second quadrant of the complex plane, we are looking for an angle $$t$$ in the second quadrant satisfying $$\tan t = \frac{b}{a} = \frac{2}{-1}=-2$$. $$\arctan ( -2 )$$ will not do, since it is in quadrant IV. We want $$\pi + \arctan ( -2 )$$, which is approximately $$2.03444$$. [The calculator must be in radian mode.]

Thus, by DeMoivre's Theorem $$( -1 + 2 i )^5$$ is approximately equal to

$$\begin{eqnarray*} 55.9 [ \cos 5(2.0344) + i \sin 5(2.0344) ]&=& 55.9(\cos10.1720 + i\sin10.1720)\\ &=& 55.9 ( -0.7336 - 0.6796 i )\\ &=& - 41.0082 - 37.9896 i. \end{eqnarray*}$$

Since the real and imaginary parts of $$( -1 + 2 i )^5$$ must be integers, the exact answer is

$$( -1 + 2 i )^5 = - 41 - 38 i$$.

## Exercise 6.3.3

Find $$( 1 + i )^7$$.

See Solution

## Roots of Complex Numbers

The inverse of finding powers of complex numbers is finding roots of complex numbers.

A complex number has two square roots, three cube roots, four fourth roots, etc. Generally, a complex number has $$n$$ nth roots. All $$n$$ of the $$n$$th roots of a complex number $$z$$ are evenly spaced around a circle centered at $$0$$ and having a radius equal to the $$n$$th root of the modulus of $$z$$. Furthermore, if $$z = | z | ( \cos t + i \sin t )$$, then one of the $$n$$ $$n$$th roots of $$z$$ is $$w_0=| z |^{1 / n} \left[ \cos \left( \frac{t}{n} \right) + i \sin \left( \frac{t}{n} \right) \right]$$. Since the n $$n$$th roots are evenly spaced around a circle, they are separated by an angle of $$\frac{2\pi}{n}$$. So, once we find the root $$| z |^{1 / n} \left[ \cos \left( \frac{t}{n} \right) + i \sin \left( \frac{t}{n} \right) \right]$$, we find the remaining $$n - 1$$ roots by adding multiples of $$\frac{2\pi}{n}$$ to $$\frac{t}{n}$$.

Example: Find the cube roots of $$1 + i$$.

First, write $$1 + i$$ in trigonometric form: $$z=1+i=\sqrt{2}\left(\cos\frac{\pi}{4}+i\,\sin\frac{\pi}{4}\right)$$

Then one root is $$w_0=\sqrt[6]{2}\left(\cos\frac{\pi}{12}+i\,\sin\frac{\pi}{12}\right)$$.

To find the arguments of the two remaining cube roots, we add multiples of $$\frac{2\pi}{3}=\frac{8\pi}{12}$$ to $$\frac{\pi}{12}$$ to get $$\frac{9\pi}{12}=\frac{3\pi}{4}$$ and $$\frac{17\pi}{12}$$.

Thus the three cube roots of $$1 + i$$ are

$$w_0=\sqrt[6]{2}\left(\cos\frac{\pi}{12}+i\,\sin\frac{\pi}{12}\right)$$

$$w_1=\sqrt[6]{2}\left(\cos\frac{3\pi}{4}+i\,\sin\frac{3\pi}{4}\right)$$

$$w_2=\sqrt[6]{2}\left(\cos\frac{17\pi}{12}+i\,\sin\frac{17\pi}{12}\right)$$

## Exercise 6.3.4

Find the four fourth roots of $$-16 i$$.

See Solution