Thursday, August 11, 2016

complex numbers - prove the following equation about inverse of tan in logarithmic for

$$\arctan(z)=\frac1{2i}\log\left(\frac{1+iz}{1-iz}\right)$$



i have tried but my answer doesn't matches to the equation .the componendo dividendo property might have been used. where
$$\arcsin(x)=\frac1i\log\left(iz+\sqrt{1-z^2}\right)$$

No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...