Finding the angle of $-2i$.

Question

Given $z = -2i$, I am to find the exponential form. Now, the radius $= 2$. The angle is derived as $\tan^{-1} \frac{y}{x}= \theta $ .

$y$ and $x$follow the form $z = x + yi$.

Now, given all this, if I was to find theta, I'd say it's $\tan^{-1} \{\frac{-2}{0}\} $. Now this doesn't make mathematical sense. However, the answer is very apparent $-\frac{\pi}{2}$. So how does the math work out in this problem?

Answer 1

Use the fact that $x=r\cos\theta,$ $y=r\sin\theta.$ The arctangent trick does not work for purely imaginary numbers, nor does it give an unambiguous result (generally two solutions).

In your specific case, then, we get $0=\cos\theta$ and $-2=2\sin\theta.$

Answer 2

$z=re^{i\theta} =r(\cos \theta +i \sin \theta)=2(0+(-1)i)$

Clearly $r=2$ and $\theta = -\frac{\pi}{2}$

For your method $\tan^{-1} \{\frac{-2}{0}\} \rightarrow -\infty $ which means $\theta= -\frac{\pi}{2}$ as $\tan \frac{\pi}{2} \rightarrow \infty $