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Hi all,

I got stuck with a problem that pop up in a paper about location of zeros for some analytic functions that I am working on.

The problem is the following: Fix two arbitrary positive integers $N$ and $n$. Let $\tilde{\ln}$ be the branch of the logarithm such that its imaginary part take values in $(0,2\pi)$. Is it true that for each $1\leq k\leq n$ the equation $$ z^N-\tilde{\ln}z-\frac{1}{N}-\frac{\ln_{\mathbb{R}}N}{N}=\frac{2\pi ik}{n}\qquad\qquad (1) $$ has exactly one solution in $$U_N(\varepsilon)=B(0,\sqrt[N]{N})- \bigcup_{j=1}^{N}B(w_j,\varepsilon)-\{0\},$$ where $w_j=\frac{1}{\sqrt[N]{N}}e^{\frac{2\pi ij}{N}}$, for all $\varepsilon<<1$ ?

Observation: In the case $N=1$, for any fixed $k$, if there are solutions for (1) it lives in the Sz√ęgo curve, which is given by the equation $|ze^{1-z}|=1$. In this case to decide if the solution is unique, for a fixed $k$, in $U_1(\varepsilon):=B(0,1)-B(1,\epsilon)-\{ 0\}$ it is enough to show that $g:U_1(\varepsilon)\to g(U(\varepsilon))$, given by $$ g(z)=\frac{e^z}{z} $$ is a biholomorphism. I suspect that the inequality $\delta(\epsilon)<|g'(z)|$ for $z\in U_1(\varepsilon)$ it is enough to prove this fact. This condition remember me the Jacobian conjecture. I don't know about this conjecture for one complex variable functions. I mentioning this case, because of the uniqueness in (1) for general $N$, can be obtaining by proving that $g:U_N(\varepsilon)\to g(U_N(\varepsilon))$ given by $$ g(z)=\frac{e^{z^N}}{z} $$ is a biholomorphism. The choice of $U_N(\varepsilon)$ it certainly not the maximal domain where this function is bijective, I am considering just based on the observation that the zeros of $g'$ are not in the closure of $U_N(\varepsilon)$.

I appreciate any help.

share|cite|improve this question
Little $n$ equals big $N$? – Marc Palm Aug 25 '11 at 22:30
Hi pm, I am most interested in the case were they are arbitrary. – Leandro Aug 25 '11 at 22:54

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