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If $f$ is a continuous function such that $g(z)=\sin{f(z)}$ is analytic, then is $f$ analytic?

I know we can take $f(z)=\bar{z}$ then $f$ is continuous but $g$ is not analytic. Same holds if we take $f(x+iy)=x$.

I tried letting $f(z)=u+iv$ then expanding $g(z)=\sin u\cosh v+i\cos u\sinh v$ taking the partial derivatives and using the Cauchy Riemann equations. That seems like a messy way to go.

coffeebelly
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  • At the points where $\sin'(f(x)) = \cos(f(x)) \neq 0$, you know that $\sin$ is locally analytically invertible, so that $f$ is analytic in a neighbourhood of these points. At the other points I am not sure at the moment. – PhoemueX Jul 16 '14 at 19:10

1 Answers1

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For points where $f(z) \neq \left(k+\frac{1}{2}\right)\pi$, the sine is locally biholomorphic, and

$$f(z) = \arcsin \left(\sin f(z)\right)$$

is holomorphic in a neighbourhood of $z$ as a composition of two holomorphic functions.

It remains to deal with the points where $f(z) = \left(k+\frac{1}{2}\right)\pi$ for some $k \in \mathbb{Z}$.

Use the identity theorem (on $\sin\circ f$, since we don't know yet that $f$ is holomorphic everywhere) to deduce that (on each component of its domain) either $f$ is constant (follows from the continuity of $f$ if $\sin \circ f$ is constant), or these points are isolated. In the latter case, the Riemann removable singularity theorem tells you that $f$ is holomorphic also in these points.

Daniel Fischer
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    Maybe you should add that the identity theorem is applied to $\sin \circ f$ and not to $f$ itself (is that so?) I was wondering for some minutes how you could apply it to $f$ when you do not yet know that $f$ is holomorphic at the points where $f(x) = (k + 1/2) \pi$. You then use (do you?) that if $\sin \circ f$ is constant on a component, then so is $f$ by continuity. – PhoemueX Jul 16 '14 at 20:04
  • Yes. We don't know yet that $f$ is holomorphic, so we don't know that we can apply the identity theorem to it. Hence the deduction is exactly as you say, @PhoemueX. – Daniel Fischer Jul 16 '14 at 20:10