Find the value of $\log \frac{a+b i}{a-b i}$

Question

Find the value of $\log \frac{a+b i}{a-b i}$

The answer is 2$i\tan^{-1}{\frac{b}{a}}$

I have calculated it upto $i \tan^{-1}{\frac{2ab}{a^{2}- b^{2}}}$

But I am unable to convert it in given answer form. Please tell me how to do it

Answer 1

Let $\theta =tan^{-1} \frac b a$. Then $tan (2\theta)=\frac {2tan(\theta)} {1-tan^{2}(\theta)}=\frac {2b/a} {1-b^{2}/a^{2}}$. Simplify.

Answer 2

Write $a+bi=re^{i\theta}$; then $$ \frac{a+bi}{a-bi}=\frac{re^{i\theta}}{re^{-i\theta}}=e^{2i\theta} $$ Thus $$ \log\frac{a+bi}{a-bi}=\frac{re^{i\theta}}{re^{-i\theta}}=\log e^{2i\theta}= 2i\theta+2k\pi i $$ It's generally wrong to say that $\theta=\arctan\frac{b}{a}$. A simple example is $$ a+bi=\frac{1}{2}-i\frac{\sqrt{3}}{2} $$ where the argument is $\theta=2\pi/3$, but $$ \arctan\frac{b}{a}=\arctan(-\sqrt{3})=-\pi/3 $$