I am trying to show that $\sin$, $\cos$, $\cosh$ and $\sinh$ only have zeros of multiplicity $1$ in the complex plane. I have found the zeros of each function but I am not sure how to show that they are of order $1$. Can somebody explain?

  • $\begingroup$ You obtained what.? $\endgroup$ – Nosrati Mar 5 '17 at 17:32

In complex analysis, the multiplicity of a root is the least $n$ such that the $n$-th derivative of the function is not $0$ at the root (cf. Wikipedia).

If you know that $z$ is a root of $sin$, so $sin(z) = 0$, then to prove it has multiplicity $1$, just show that $sin'(z) = cos(z) \neq 0$.

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