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Let, $a\in \mathbb R$ be fixed. Find the set of $z\in \mathbb C$ for which $$\sum_{n=1}^{\infty}n^{i(z^2+a)}$$ represents an analytic function.

I know that if the radius of convergence of a power series about any point $p$ is $R>0$ then the function is analytic in the neighbourhood of $p$ & if $R=\infty$ then the function is an entire function.

But, from this series I could not conclude anything...

Please help..

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  • $\begingroup$ I feel thinking in terms of power series maynot help. I tried with $a=0$ to get that $f(z)=\sum cos (c_nz^2)+i sin(c_nz^2)$ where $c_n=log_e(n)$ . But I don't know how to proceed further. $\endgroup$
    – Srinivas K
    Mar 12, 2015 at 7:35

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$$\sum_{n=1}^{\infty}n^{i(z^2+a)}$$ is the generalized harmonic series calculated in $-i(z^2+a)$. The generalized harmonic series (in wikipedia: p-series or hyperarmonic series: http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29#P-series ) $$\sum_{n=1}^{\infty}n^{-s}$$ converges to $\zeta(s)$ ( i.e. Riemann's Zeta that is an analytic function in $\mathbb C$ except for $s=1$, where a pole is present) when $Re(s)>1$, and it diverges otherwise.

Therefore your series should converge for $$Re(-i(z^2+a))>1$$ But $a$ is real therefore: $$Re(-i(z^2))>1$$ $$Re(2xy-i(x^2-y^2))>1$$

Then we have $$2xy>1$$ is the region of convergence of your series. In such region your series coincides with $\zeta(-i(z^2+a))$, therefore it represents an analytic function.

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