Consider the equations$$ 1+\frac{1}{z^k}=0 \quad\mbox{and}\quad 1+\frac{1}{z^k}+\frac{1}{(z+1)^k}=0,$$where $k$ is a positive integer $\ge 4$. I would like to show for instance that the number of zeros in $\frac{\pi}{2}\lt \mbox{Arg}(z) \lt \frac{2\pi}{3}$ is the same for both equations. I'm not sure if this is true, I'm basing this on numerical evidence.
Of course the first equation is equivalent to $z^k=-1$, so all the zeros lie on the unit circle and it is easy to explicitely compute the zeros. The problem is with the second equation, there can be some zeros outside the unit circle, so the condition $\frac{\pi}{2}\lt \mbox{Arg}(z) \lt \frac{2\pi}{3}$ is really important here. Intuitively what I think happens is that in this region the term $1/(z+1)^k$ is small for big $k$, but when $\mbox{Arg}(z)$ is close to $2\pi/3$ this is not always the case (in that case $|1/(z+1)^k|\approx 1$, so it does contribute to the second equation).
Is there a way to get past the issue near $2\pi/3$ and show that the two equations have the same number of zeros and moreover the solutions are close to each other? (Again this is from numerical evidence)