Prove there is an entire function $f$ such that for any branch $g$ of $\sqrt{z}$, $$\sin^2(g(z))=f(z)$$ for all $z$ in the domain of definition of $g$.

I don't know how to overcome the fact that $g$ is not defined everywhere on the complex plane. Thanks for any help.

  • $\begingroup$ The following fact is useful for your purpose >A function is analytic at a point $z_0$ iff it possess a power series at $z_0$. $\endgroup$ Jan 3 '15 at 1:44
  • 1
    $\begingroup$ The relation $\sin^2(z) = \frac{1}{2} - \frac{1}{2}\cos(2z)$ togeather with $\cos(2z) = \sum_{k=0}^\infty \frac{(2z)^{2k}(-1)^{k}}{(2k)!}$ might come in handy. $\endgroup$
    – Winther
    Jan 3 '15 at 2:13
  • $\begingroup$ How about those points where $g$ is not defined? Is $f$ still good there? $\endgroup$
    – Liu
    Jan 3 '15 at 2:51
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    $\begingroup$ We can define an entire $f$ that agree with $\sin^2(g(z))$ for all $z$ where the latter is defined. On points where $g$ is not defined $\sin^2(g(z))$ does not make sense, but $f$ will still be analytic there (as the power-series converges everywhere). $\endgroup$
    – Winther
    Jan 3 '15 at 2:55
  • $\begingroup$ Thank you so much for explaining it to me so clearly. $\endgroup$
    – Liu
    Jan 3 '15 at 3:21

We have $$\sin^2(z) = \frac{1-\cos(2z)}{2}$$

$$\cos(z) = \sum_{k=0}^\infty \frac{z^{2k}(-1)^k}{(2k)!}$$

which gives

$$\sin^2(g(z)) = \frac{1}{2}\sum_{k=1}^\infty \frac{(g^2(z))^k 4^k(-1)^{k+1}}{(2k)!} = \frac{1}{2}\sum_{k=1}^\infty \frac{z^k 4^k(-1)^{k+1}}{(2k)!}$$

except for points on the branch-cut $z\in (-\infty,0]$ where $g$ is not defined, since for any branch $g(z)$ of $\sqrt{z}$ we have $g^2(z) = z$.

Now it's easy to check that the power-series

$$f(z) \equiv \frac{1}{2}\sum_{k=1}^\infty \frac{z^k 4^k(-1)^{k+1}}{(2k)!}$$

converges everywhere in the complex plane any therefore representes an entire function which equals $\sin^2(g(z))$ for all points where the latter is defined.


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