1. ## Stoke's Phenomena

$\displaystyle \delta>0, \ \ show \ \ cosh(z) \approx \frac{1}{2}e^{z}, \ \ |arg(z)|<\frac{\pi}{2}-\delta$

and

$\displaystyle cosh(z)\approx \frac{1}{2}e^{-z}, \ \ \frac{\pi}{2}+\delta<arg(z)<\frac{3\pi}{2}-\delta$

not need to go along rays.

2. For instance, for the first one: $\displaystyle \cosh z=\frac{e^z}{2}(1+e^{-2z})$ and $\displaystyle |e^{-2z}|=e^{-2\Re(z)}=e^{-2|z|\cos({\rm arg}(z))}$ $\displaystyle \leq e^{-2|z|\cos(\frac{\pi}{2}-\delta)}=e^{-2|z|\sin\delta}$ for all $\displaystyle z$ such that $\displaystyle |{\rm arg}(z)|<\frac{\pi}{2}-\delta$.

This proves that $\displaystyle \cosh z = \frac{e^z}{2}(1+o(1))$ as $\displaystyle |z|\to\infty$ and $\displaystyle |{\rm arg}(z)|<\frac{\pi}{2}-\delta$, uniformly for $\displaystyle z$ in this domain.

3. ## wat

what about the case $\displaystyle \delta=0$

4. Originally Posted by pengchao1024
what about the case $\displaystyle \delta=0$
Then this is false. If you choose $\displaystyle |z|\to\infty$ that gets very close to the imaginary axis, $\displaystyle \cosh z$ gets close to cosine of a real number, hence it can be bounded ($\displaystyle \cosh(ix)=\cos(x)$).