# 14.3 The trigonometric form of a complex number

When we say $z:=x+iy$ (also known as \sayz is defined as $x+iy$), $x$ and $y$ can be anything! For example, $x$ and $y$ could equal $\cos(\theta)$ and $\sin(\theta)$ respectively. By setting different values of theta, we can obtain any point on a circle with radius 1, and centre $(0,0)$. For example, if we have $z:=\cos(\theta)+i\sin(\theta)$, then to obtain the number $1$, we can just set $\theta=0$. To be able to obtain every complex number, however, we need to introduce a second variable (which we can call $r$). This \sayscales the circle, so that for each value of $r$, the values of $\theta$ between $0$ and $2\pi$ ($2\pi$ is not inclusive) we can obtain a circle with that radius. Overall, we can write any complex number in the form

$z:=r\left(\cos(\theta)+i\sin(\theta)\right).$ (14.36)

Additionally, we can do this with $r\geqq 0$ and $-\pi\leqq\theta\leqq\pi$.

To find the trigonometric form of a given complex number there are two ways.

To convert a complex number (e.g. $1+3i$) into trigonometric form, the first way algebra is to use algebra (in particular \saycomparing coefficients). By comparing $r\cos(\theta)+ri\sin(\theta)$ with $1+3i$, we obtain that $r\cos(\theta)=1$ and that $r\sin(\theta)=3$. Thus

 $\displaystyle r^{2}\cos^{2}(\theta)+r^{2}\sin^{2}(\theta)$ $\displaystyle=r^{2}(\cos^{2}(\theta)+\sin^{2}(\theta)$ $\displaystyle=r^{2}$ $\displaystyle=2\text{ (from 1+3i)}$

from which we can deduce that $r=\pm\sqrt{2}$. The other thing we can do is divide through, thus obtaining that $\frac{r\sin(\theta)}{r\cos(\theta)}=\frac{1}{1}$, and thus that $\theta=\arctan(1)$ (which is $\frac{\sqrt{2}}{2}$). Overall, we can then write that

$1+i=\sqrt{2}e^{\frac{\sqrt{2}}{2}i}$ (14.37)

which we can also do for any complex number.

The second way involves geometry (I still need to find where I originally wrote my notes on this, but the method is to draw the complex number - not necessary, but usually helpful - and to then find the modulus and argument of the complex number). First we can apply this useful fact:

###### Theorem 14.3.1

For a complex number $z=r(\cos(\theta)+i\sin(\theta))$, $\arg(z)=\theta$ and $|z|=r$

The proof is definitely not relevant for any A Level examination, but it’s also not too difficult.

• First we prove that $\arg(z)=\theta$. We know that for a complex number $x+iy$, there are several cases for the argument (rather unfortunately we need to consider each quadrant separately)

• In the first case, $x,y\geqq 0$, where $\arg(z)=\arctan\left(\frac{y}{x}\right)$. For our $z$, this means

 $\displaystyle\arg(z)$ $\displaystyle=\arctan\left(\frac{r\sin(\theta)}{r\cos(\theta)}\right)$ (14.38) $\displaystyle=\arctan(\tan(\theta))$ (14.39) $\displaystyle=\theta$ (14.40)
• TODO: go through the other three cases

• Second we prove that $|z|=r$. This proof is quite a bit more satisfying than the previous proof, we just apply the definition of the modulus, that is that

 $\displaystyle z^{2}$ $\displaystyle=\sqrt{x^{2}+y^{2}}$ (14.41) $\displaystyle=\sqrt{\Big{(}\big{(}r\cos(\theta)\big{)}^{2}+\big{(}r\sin(\theta% )\big{)}^{2}\Big{)}}$ (14.42) $\displaystyle=\sqrt{r^{2}\big{(}\cos^{2}(\theta)+\sin^{2}(\theta)\big{)}}$ (14.43) $\displaystyle=\sqrt{r^{2}}$ (14.44)

Note that we also defined $r\geqq 0$ (well, hopefully we did) and thus

$\sqrt{r^{2}}=r$ (14.45)

## 14.3.1 Using the trigonometric form of a complex number

###### Example 14.3.1

This is probably not the most exciting example, but we might want to consider the case of

$z=2(\cos(330^{\circ})-\sin(330^{\circ}))$ (14.46)

If we can write this in trigonometric form, then by 14.3.1 we can just read the argument and modulus. Examining the expression, it looks pretty close to something in the form

$z=r(\cos(\theta)+\sin(\theta)).$ (14.47)

The only problem we have is that there’s a pesky negative hanging around in there, but not to worry, we can apply two useful facts here; $\cos(\theta)=\cos(-\theta)$ and $\sin(\theta)=-\sin(-\theta)$ (these two equations are referred to as the \sayeven and \sayodd property of $\cos$ and $\sin$ respectively - see Section 14.7.3 for a proof of this).

Therefore, we can rewrite $z$

 $\displaystyle z$ $\displaystyle=2(\cos(330^{\circ})-\sin(330^{\circ}))$ (14.48) $\displaystyle=2(\cos(-330^{\circ})+\sin(-330^{\circ}))$ (14.49) $\displaystyle=2(\cos(-30^{\circ})+\sin(-30^{\circ}))$ (14.50)

Therefore the argument is $-30^{\circ}=-\frac{\pi}{6}\text{ rad}$