Exponential type of $\sin z$

Question

An entire function $f$ is of exponential type if $\,\lvert\, f(z)\rvert\le C\mathrm{e}^{\tau\lvert z\rvert},\,$ for all sufficiently large values of $\lvert z\rvert$.

The exponential type of $f$ is the infimum of all $\tau$ satisfying the above inequality.

I am trying to determine the exponential type of $\sin z$, so far without any luck - hope you guys can help me!

Answer 1

First observe that ($z=x+iy$) $$ \lvert\sin z\rvert\le\frac{1}{2}\lvert\mathrm{e}^{iz}\rvert+\frac{1}{2}\lvert\mathrm{e}^{-iz}\rvert=\frac{1}{2}\big(\mathrm{e}^{y}+\mathrm{e}^{-y}\big)\le \frac{1}{2}\big(\mathrm{e}^{\lvert z\rvert}+\mathrm{e}^{\lvert z\rvert}\big)=\mathrm{e}^{\lvert z\rvert}. $$ Hence $\tau\le 1$.

Then, for $z=iy$, with $y>0$, we have $$ \lvert\sin z\rvert\ge\frac{1}{2}\big(\mathrm{e}^y-\mathrm{e}^{-y}\big)\ge \frac{1}{4}\mathrm{e}^y=\frac{1}{4}\mathrm{e}^{\lvert z\rvert}. $$ The second inequality holds for $\mathrm{e}^{2y}\ge 2$ or $y\ge\frac{\log 2}{2}$.

Hence $\tau\ge 1$.