# Show that $\text{sin}(\bar z)$ is not holomorphic using uniqueness theorem.

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?

## 2 Answers

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.

• Thanks feeling so dumb for entirely missing why the theorem is actually so useful, hah. – Dole Dec 5 '18 at 5:57

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 $$z=i$$.