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My professor gave us this problem.

Find all complex numbers $z\in \mathbb C$ such that $$\frac{1}{1^z}+\frac{1}{3^z}+\frac{1}{5^z}+\cdots=\frac{1}{2^z}+\frac{1}{4^z}+\frac{1}{6^z}+\cdots$$

I removed my try because it's wrong.

Did
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Superbus
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    I do not think that "every term of the series is strictly positive", because there is no "positive" when it comes to complex numbers, in some sense. – Yiyuan Lee Apr 05 '14 at 08:40
  • Oh yes, good point.. – Superbus Apr 05 '14 at 08:44
  • @YiyuanLee But still, the dirichlet series is very well studied. Does it equal $0$ for any $n$? – Guy Apr 05 '14 at 08:45
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    @YiyuanLee So the problem isn't meant to be solved? Or is the professor trying to get his students to solve it? =) – Guy Apr 05 '14 at 08:50
  • Seriously though, I think $\Re(n)\gt1$ is implied here. I think we can ignore equating this to $0$ by analytic continuation. For the original equation, $n=-2$ is certainly not a valid answer, yet $\zeta(-2)=0$ – Guy Apr 05 '14 at 08:51
  • Yes, that is implied. Both sides need to converge, and be equal, so I don't think the riemann hypothesis is totally relevant! – Superbus Apr 05 '14 at 08:55
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    Answer here: http://en.wikipedia.org/wiki/Dirichlet_eta_function – Jason Apr 05 '14 at 08:55
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    @Jason No. The Dirichlet eta function, accounts for analytic continuation as well. $D(-2)=0$, but here both sides must actually converge – Guy Apr 05 '14 at 09:00

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The sum of both sides is clearly $\zeta (z)$ by definition. I suppose you also know that
$\zeta (z)= \prod _k^{\infty } \frac{1}{1-\frac{1}{\left(p_k\right){}^z}}$
Now, if we start the product not from p1=2 but from p2=3 then we generate all odd integers (no factor 2 available anymore). By difference we get the right equation. We get :
$2^{-z} \left(2^z-1\right) \zeta (z) = 2^{-z} \zeta (z)$ or, equivalently $2^{-z}=2^{-z} \left(2^z-1\right)$
Solution(s) : $z = 1+\frac{2 i \pi n}{\log (2)}$ for all integer n.
Say z=1. Both left and right side go to infinity, but get to be equal for z approaching 1 from above.

Wouter M.
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