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Consider the complex function $$f(z)=\sum_{k=1}^n z^{\alpha_k}\,, \quad z\in\mathbb C,\;\Re z>0$$ where $\alpha_k$ are real numbers (assume positive without loss of generality).

What can I say about the number of zeros of $f$? Can I bound it from above? If the $\alpha_k$ were positive integers, then $f$ would have at most $N=\max_k \alpha_k$ zeros on $\Re z>0$...

Can I found a lower bound for $|\Im z|$, when $z$ is a zero of $f$ and $\Re z>0$?

tituf
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  • For non-integer $\alpha$, $z^\alpha = e^{\alpha \log z}$ does not have a branch which is holomorphic in $\Bbb C$, so you have to clarify first how $f$ is defined in that case. – Martin R Dec 22 '15 at 13:56
  • Right, assume that $\Re z>0$, so that a brach of the logarithm is well-defined – tituf Dec 22 '15 at 13:58
  • I think this has been asked twice before, but for the life of me I can't find the questions. – Antonio Vargas Dec 22 '15 at 14:07
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    What about this one: Number of roots of $x^a-1=0$ with $a \in \mathbb{C}$ ? For irrational $a$ it has infinitely many solutions. – Martin R Dec 22 '15 at 14:16
  • Thank you. So if $\alpha_k$ is irrational, I cannot even hope to bound the zeros far from the positive real axis? I mean, can I found a lower bound for $|\Im z|$ when $z$ is a zero of $f$? – tituf Dec 22 '15 at 14:28
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    According to the referenced answer, the solutions of that equation are dense on the unit circle, so: No. With three or more terms things probably get much more complicated. – If you substitute $w = \log z$ then you get an exponential polynomial in $w$. I am not an expert on this topic, but as far as I remember, the zeros of E.P.s can be described asymptotically. A classical reference is http://www.ams.org/journals/tran/1929-031-04/S0002-9947-1929-1501506-6/S0002-9947-1929-1501506-6.pdf. – Martin R Dec 22 '15 at 14:37

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