Question

Prove that `log((a+ib)/(a-ib))=2itan^(-1)  b/a      &    cos[ilog((a+ib)/(a-ib))=(a^2-b^2)/(a^2+b^2)]`

Answer 1

let L`=log((a+ib)/(a-ib))`

Using logarithmic properties ,

L` =log(a+ib)-log(a-ib)`

`=1/2log(a^2+b^2)+itan^(-1)  b/a-[1/2log(a^2+b^2)-itan^(-1)  b/a]`

L`=2itan^(-1)  b/a`

`therefore log((a+ib)/(a-ib))=2itan^(-1)  b/a` 
Hence Proved.

`therefore (a+ib)/(a-ib)=e^(2itan^(-1)  b/a)=cos(2tan^(-1)  b/a)+isin(2tan^(-1)  b/a)`

`therefore (a+ib)/(a-ib)xx(a+ib)/(a+ib)=cos(2tan^(-1)  b/a)+isin(2tan^(-1)  b/a)`

`(a^2-b^2)/(a^2+b^2)+"imaginary"=cos(2tan^(-1)  b/a)+isin(2tan^(-1)  b/a)`

Separate real and imaginary parts

`cos(2tan^(-1)  b/a)=(a^2-b^2)/(a^2+b^2)`

From 1^st result ,

`cos[ilog((a+ib)/(a-ib))]=(a^2-b^2)/(a^2+b^2)`

Hence Proved.