# Finding the polar form of a complex number

I have the following complex numbers : -3,18 +4,19i

I can calculate $r=\sqrt{a^2+b^2}$

Which gives r=5,26

now I know that cos $\theta = \frac{a}{r}$

gives $\theta=127,20$ degrees

But when I do the same with : $sin \theta = \frac{b}{r}$

it gives me $\theta= 52,80$ degrees

Why does it do this ? I know that it must give 127,20 degrees for both of them...

If you draw your number in the complex plane you see that $\theta = 127.2$ could be correct but that $\theta = 52.8$ lies is in a different qudarant, so that's definitely not the right answer.
I expect that you solved the equation $\sin \theta = \frac{b}{r}$ using the inverse sine function. When working with the inverse trig functions one has to remember that their range is limited (otherwise they wouldn't meet the definition of a function (each input has a single output)). Specifically:
• Inverse sine $= \sin^{-1}$ has domain $[-1, 1]$ and range $[-\frac{1}{2}\pi, \frac{1}{2}\pi]$
• Inverse cosine $= \cos^{-1}$ has domain $[-1, 1]$ and range $[0,\pi]$
• Inverse tangent $= tan^{-1}$ has domain $(-\infty, \infty)$ and range $(-\frac{1}{2}\pi, \frac{1}{2}\pi)$.