Sharpest upper bound for $\lvert\cos(z)\rvert$

Question

Is it possible to obtain a sharper estimate for $\lvert\cos(z)\rvert$ than $$ \lvert\cos(z)\rvert \leq e^{\lvert y\vert},$$ for every $z=x+iy?$

Answer 1

$$\left\|\cos(x+iy)\right\|^2 = \cos(x+iy)\cos(x-iy)=\frac{1}{2}\left[\cos(2x)+\cosh(2y)\right] $$ hence the sharpest possible bound (in terms of $y$ only) is $\left\|\cos(z)\right\|\leq \sqrt{\frac{e^{2|y|}+1}{2}}.$

Answer 2

An easy improvement of your inequality is to observe that $$\lvert\cos z\rvert = \lvert(\mathrm{e}^{\mathrm{i} z} + \mathrm{e}^{-\mathrm{i} z}) \mathbin{/} 2\rvert \leq (\lvert \mathrm{e}^{\mathrm{i} z}\rvert + \lvert \mathrm{e}^{-\mathrm{i} z}\rvert) \mathbin{/} 2 = (\mathrm{e}^{-y} + \mathrm{e}^{y}) \mathbin{/} 2 = \mathop{\mathrm{cosh}} y\text.$$ Moreover, the inequality is sharp as soon as $\mathrm{e}^{\mathrm{i} z}$ and $\mathrm{e}^{-\mathrm{i} z}$ are positively colinear, which occurs whenever $x$ is a multiple of $\pi$.