From complex to polar

Question

In a scientific paper, the authors express the following eigenvector in the polar form $$\frac{1}{\sqrt{2}}\left\{0,\frac{i \sqrt{a^2+b^2}}{a+i b},1,0\right\}$$ to take the form: $$\frac{1}{\sqrt{2}}\left\{0,e^{i \theta },1,0\right\}$$ where $\theta =\tan ^{-1}\left(\frac{a}{b}\right)$. Can I know how they did this?

Answer 1

If $\tan\theta=a/b$, then

$$ \sin\theta=\frac{a}{a^2+b^2}\\ \cos\theta=\frac{b}{a^2+b^2} $$

Now,

$$i\frac{\sqrt{a^2+b^2}}{a+ib}=i\frac{\sqrt{a^2+b^2}(a-ib)}{a^2+b^2}\\ =i\frac{a-ib}{\sqrt{a^2+b^2}}\\ =\cos\theta+i\sin\theta=e^{i\theta} $$