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

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

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

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:

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

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  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)

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    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)

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    TODO: go through the other three cases

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  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)