If $\tan(\alpha+i\beta)=(x+iy)$ then prove that:

Question

If $\tan(\alpha+i\beta)=(x+iy)$ then prove that:

• $x^2+y^2+2x\cot2{\alpha}=1$
• $x^2+y^2-2y\coth2{\beta}=1$

I think that using the concept of logarithm will helpful to solve the problem but I am not able to do so. A detail solution explaining the concept will be helpful.

Thanks in advance !

NB: After going through all the suggestion questions I think all my doubts are cleared with this : $\tan (x+yi)=\alpha +\beta i \Rightarrow \tan(x-yi)=\alpha-\beta i$?

Answer 1

Use $$\tan(2\alpha)=\tan\{\alpha+i\beta+(\alpha-i\beta)\}=?$$

and $$\tan(2i\beta)=\tan\{\alpha+i\beta-(\alpha-i\beta)\}=?$$

Now $\tan(2i\beta)=i\tanh(2\beta)$