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$\zeta$ function has only one pole at $z=1$. It also has order $1$. If $\zeta$ omits $c\in C$ then $g:=1/(\zeta-c)$ is entire with one simple zero at $1$. As it is of order $1$, it must be $g(z)=(z-1) …

There is a truly elementary proof. Nothing but high school mathematics + the notion of limit is used. First one proves Cauchy's inequality for polynomials: $$|f(0)|\leq M(r),$$ where $M(r)$ is the ma …

Linear functional equations can be solved with Fourier transform. Let $\lambda_k$ be the roots of the equation $e^\lambda-1=\lambda$. There are infinitely many such roots. Then $$f(z)=\sum_k a_ke^{\ …

The earliest application is the Mercator projection which was introduced long before the complex exponential was defined in the way we define it nowadays. $z\mapsto e^z$ is considered as a map from th …

This is a classical problem, but only partial results are available: MR3155684 Hayman, W. K.; Tyler, T. F.; White, D. J. The Blumenthal conjecture, in Complex Analysis and Dynamical Systems V, 149–157 …

Such a function does not exist. Assume first that $f$ is real on the real line. We use the theorem of Marchenko–Ostrovski that if $f$ is entire and all zeros and solutions of $f(z)=3$ are real, then $ …

This is a special case of the main theorem in the paper by I. N. Baker, J. A. Deddens, and J. L. Ullman, A theorem on entire functions with applications to Toeplitz operators, Duke Math. J. Volume 41, …

Yes, there are such functions. Take a very narrow region $D$ containing the positive ray, with nice boundary and such that $D$ intersects every any horizontal line other than the real line by a bounde …

There are many good books, but the choice depends on your background and on your needs and on your taste. For what purpose do you study complex variables? Do you like geometry or formulas? If your aim …

Here is a simple proof for complex-analytic case. If restrictions of $f$ on all complex lines are analytic, then $f$ is analytic. This reduces the problem to the case $n=1$. Now $f^2$ is analytic so n …

This was conjectured by J. Borcea, and a counterexample was constructed by J. Langley: MR2317957 Langley, J. K. Equilibrium points of logarithmic potentials on convex domains, Proc. Amer. Math. Soc. …

$2n$ is incorrect. The correct upper estimate is $n^2$ (if the number is finite). Indeed, let $P(z,\overline{z})$ be a polynomial of degree $n$. Writing $z=x+iy$ and $\overline{z}=x-iy$ we obtain one …

I don't have a complete proof yet, but I have a plausible conjecture. Let $\mu$ be a probability measure in the plane, define the potential $$u(z)=\int\log|1-z/t|d\mu(t).$$ Then I conjecture that $\mu …

A simpler proof can be obtained as follows. Proving by contradiction, suppose it has no zeros. Since this is an entire function of order one, it must be $\exp(az+b)$. So we have the identity $$\sum_{k …

There are very many different proofs of Picard's theorem, and some of them are really "simple". (Picard's original proof occupies about 2 lines, using the things already known at that time). None of t …