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While studying for a test I have encountered this problem,

Is there a holomorphic function $g:\mathbb{C}\rightarrow\mathbb{C}$ such that $g(z(\pi-z))=\sin(z)$, $(\forall z\in \mathbb{C})$?

I thought of developing a Taylor series for $\sin(z)$ and $g(w)$ around $w,z=0$, saying coefficients must be equal for same degree variable $z$.

It doesn't seem to work for me though... Any help would be appreciated :)

Martin R
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Daniel
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1 Answers1

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Note that $$ z(\pi - z) = \frac {\pi^2}{4} - \left(z - \frac{\pi}{2}\right)^2 $$ so with the substitution $z - \frac{\pi}{2}= w$ your identity is equivalent to $$ g\left(\frac {\pi^2}{4} - w^2\right) = \cos(w) $$

Now use the fact that the cosine is an even function, therefore $\cos(w) = h(w^2)$ for some entire function $h$. Then $g : \Bbb C \to \Bbb C$ defined by $$ g(u) := h\left(\frac{\pi^2}{4} - u\right) $$ is the desired solution.

José Carlos Santos
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Martin R
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  • Thank Martin, But why can you assume existence of such a function $h$? How does the fact that cosine is even help you? – Daniel Apr 02 '16 at 16:17
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    @Daniel: You can (for example) use the fact that the power series of the cosine has only even coefficients. See the final part of http://math.stackexchange.com/a/1177343/42969 for an explicit construction. – Martin R Apr 02 '16 at 16:20
  • Many thanks :)) – Daniel Apr 02 '16 at 16:33