# Location of solutions in complex analysis

Try to show that the equation $$(z-1)^ne^z = a$$ where $n$ is a positive integer and $|a|<1$ has exactly $n$ solutions in the right half plane.

I have tried to take logs of both sides and then apply Rouche's theorem, but I haven't been able to finish the problem this way.

• To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help you if you show that you've tried the problem yourself. Apr 24, 2013 at 20:37
• I have tried to take logs of both sides and then apply Rouche's theorem. Apr 24, 2013 at 20:42
• @EagerStudent Taking logs of complex numbers is a very risky business since logarithms are multi-values finctions. For example: $\ln z = \ln|z| + i \operatorname{arg} z$. For example, $\ln 1 \in \{ 2\pi i n : n \in \mathbb{Z}\}$. Apr 24, 2013 at 20:47

$$f(z):=(z-1)^n\;,\;\;g(z):=-\frac{a}{e^z}$$
Both are holomorphic functions, and on the boundary of $\,\{z\in\Bbb C\;;\;|z-1|\le 1\}\,$ we have
$$|g(z)|=\frac{|a|}{e^{\text{Re}(z)}}\le|a|<1=|z-1|=|f(z)|\implies$$
By Rouche's theorem, $\,f(z)\;\;\text{and}\;\;\;f(z)+g(z)\;$ have the same number of zeros inside the domain $\,|z-1|<1\,$ ...