Proof that $\arctan z+\arctan \frac{1}{z}$ does not depend on $\operatorname{Im}z$ when $\operatorname{Re}z\ne 0$ and $\operatorname{Im}z\ne 0$

Question

I proved that $$\begin{align*}\arctan z+\arctan \frac{1}{z}&=\frac{\pi}{2} \, \text{if} \, z\gt 0\\&=-\frac{\pi}{2} \, \text{if} \, z\lt 0\end{align*}$$ using the fact that $$\frac{d}{dz}\left(\arctan z+\arctan \frac{1}{z}\right)=0, \, \operatorname{Re}z\ne 0, \, \operatorname{Im}z\ne 0.$$ I suspect that $$\begin{align*}\arctan z+\arctan \frac{1}{z}&=\frac{\pi}{2} \, \text{if} \, \operatorname{Re}z\gt 0\\&=-\frac{\pi}{2} \, \text{if} \, \operatorname{Re}z\lt 0.\end{align*}$$ Therefore, when $\operatorname{Re}z\ne 0$ and $\operatorname{Im}z\ne 0$, $\arctan z+\arctan \frac{1}{z}$ does not depend on $\operatorname{Im}z$. If this is true, how can I prove it?

Answer 1

Use the logarithmic form of $\tan^{-1}$: $$\tan^{-1}z+\tan^{-1}\frac1z=\frac i2\left(\ln\frac{i+z}{i-z}+\ln\frac{i+1/z}{i-1/z}\right)$$ $$=\frac i2\ln\frac{(i+z)(i+1/z)}{(i-z)(i-1/z)}=\frac i2\ln\frac{-1+zi+i/z+1}{-1-zi-i/z+1}$$ $$=\frac i2\ln\frac{zi+i/z}{-(zi+i/z)}=\frac i2\ln-1$$ The reason two values are obtained for the original expression is because of the branch cuts used in defining $\tan^{-1}$, and this is reflected in the $\ln-1$ at the end, which has principal value $i\pi$ but may also be taken as $-i\pi$, producing the two different results $\pm\frac\pi2$. In particular, the sign of the real part of both $\tan^{-1}z$ and $\tan^{-1}\frac1z$ is the same as the sign of the real part of $z$ or $\frac1z$ respectively with the usual definition of $\tan^{-1}$ in programming languages.