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I want to show that $\text{sin}(\bar z)$ is not analytic using the uniqueness theorem.

The theorem essentially states that if we have a series $z_n$ such that non-constant $f(z_n)$ is zero for each $n$, then the function is not holomorphic if the infinite limit exists, but is not equal to any $z_n$.

The problem is $\text{sin}(\bar z)$ has zeros at $z=n\pi$. The theorem is directly of no help. What transform should be performed?

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Let $f(z)=\text{sin}(\bar z)$ and $g(z)=\sin(z)$. Suppose that $f$ is holomorphic.

We have $f(z)=g(z)$ for real $z$. Hence, by the uniqueness theorem, $f(z)=g(z)$ for all $z$. But this is absurd.

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  • $\begingroup$ Thanks feeling so dumb for entirely missing why the theorem is actually so useful, hah. $\endgroup$ – Dole Dec 5 '18 at 5:57
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If $\sin (\overline {z})$ is holomorphic then it must coincide everywhere with $\sin \, z$ because these two holomorphic functions are equal on the real line (which has limit points). This is a contradiction because these functions are not equal when $z=i$.

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