Q:

Show $\lvert\sin z\rvert \leq C \lvert z\rvert$ $\forall \lvert z\rvert\leq 1$

Show $\lvert\sin z\rvert \leq C \lvert z\rvert$ $\forall \lvert z\rvert\leq 1$

The question asks me to show the constant $C$ exists and to estimate it.
I only know $\sin z = \dfrac{e^{iz}-e^{-iz}}{2i}$, so I want to find $C$ such that $\lvert e^{iz} - e^{-iz}\rvert \leq C \lvert z\rvert$ for $\lvert z\rvert \leq 1$, but I do not know how to solve it, thank you for helping.

A:

Hint: $|e^{iz}| \le e^{|z|}$ so $|\sin(z)| \le e$ for $|z|=1$.

A:

Note that $\sin' = \cos$ and $|\cos z| \le \cosh (\operatorname{im} z) \le \cosh 1$ for $|z| \le 1$. Hence $|\sin z| \le (\cosh 1) |z|$.

