$$\arctan(z)=\frac1{2i}\log\left(\frac{1+iz}{1-iz}\right)$$
i have tried but my answer doesn't matches to the equation .the componendo dividendo property might have been used. where
$$\arcsin(x)=\frac1i\log\left(iz+\sqrt{1-z^2}\right)$$
$$\arctan(z)=\frac1{2i}\log\left(\frac{1+iz}{1-iz}\right)$$
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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