How to solve complex equations of the form $(z+a)^n=(z+b)^m$? I know how to solve complex equations of the form $(z+a)^n=r$ but how does the solution change when we have an equation of the form $(z+a)^n=(z+b)^m$?
Suppose we may represent $z+a=re^{i\phi}$ and $z+b=se^{i\psi}$, where $r, s, \phi, \psi$ can be calculated.
Then, our equation becomes $$r^ne^{in\phi}=s^me^{im\psi}$$
Equating like terms (we must, since the equation must be consistent in modulus, and the $e$ part has modulus $1$), we see:
$$r^n=s^m\implies \frac mn=\frac{\ln r}{\ln s}$$
$$n\phi+2k\pi=m\psi \implies \frac mn=\frac\phi\psi+\frac{2k}{n\psi\pi}$$
for some $k\in \Bbb Z$
We can solve this system for
$$n=\frac{2k\ln[s]}{\pi\ln[r^{\psi}s^{-\phi}]}\implies m=\frac{2k\ln[r]}{\pi\ln[r^{\psi}s^{-\phi}]}$$ and find a solution $(m, n)$ for each $k$
Suppose we may represent $z+a=re^{i\phi}$ and $z+b=se^{i\psi}$, where $r, s, \phi, \psi$ can be calculated.
Then, our equation becomes $$r^ne^{in\phi}=s^me^{im\psi}$$
Equating like terms (we must, since the equation must be consistent in modulus, and the $e$ part has modulus $1$), we see:
$$r^n=s^m\implies \frac mn=\frac{\ln r}{\ln s}$$
$$n\phi+2k\pi=m\psi \implies \frac mn=\frac\phi\psi+\frac{2k}{n\psi\pi}$$
for some $k\in \Bbb Z$
We can solve this system for
$$n=\frac{2k\ln[s]}{\pi\ln[r^{\psi}s^{-\phi}]}\implies m=\frac{2k\ln[r]}{\pi\ln[r^{\psi}s^{-\phi}]}$$ and find a solution $(m, n)$ for each $k$