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Let $f\left(z\right)$ be an analytic function on an open disk $D$ (of possibly infinite radius) centered at zero satisfying the functional equation:

$$f\left(z^{3}\right)=f\left(-z^{3}\right),\textrm{ }\forall z\in D$$

Can I conclude that $f\left(z\right)=f\left(-z\right)$ for all $z$ in some open neighborhood of the origin?

This isn't for homework. I simply want—and desperately need—an answer, and an explanation why, so that I can hop over this puddle and continue with the body of my research.

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I'll discuss one interesting case:

Could we say: $\forall z \in D$, $\exists k \in D$, s.t. $k^3 = z$? If so $f(z) = f(k^3) = f(-k^3) =f(-z)$.

Write $z=|z|e^{i\theta} \in D$, then could we find $k=z^{\frac{1}{3}} = |z|^{\frac{1}{3}}e^{\frac{i\theta}{3}}\in D$ for any $z\in D$? The answer is NO, because if you take the radius of $D$ to be smaller than $1$, and take $z$ to be close to the boundary of $D$, then $|z|^{\frac{1}{3}}$ will be larger than $z$ and fall outside of $D$.

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  • $\begingroup$ What if the radius of convergence of the series is infinite? I.e., what if $f$ is an entire function? Can I conclude that $f$ is even? $\endgroup$ – MCS May 6 '17 at 1:46
  • $\begingroup$ @MCS if $D=\mathbb C$, then yes! This is because $\forall z \in \mathbb C$, it is true that you'll always find $k \in \mathbb C$, s.t. $k^3=z$ $\endgroup$ – Yujie Zha May 6 '17 at 2:02

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