Tuesday, August 25, 2015

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 $$\arcsin(x)=\frac1i\log\left(iz+\sqrt{1-z^2}\right)$$

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