Let $z=Re^{i \alpha}$ $\alpha$ is given number and $\alpha \in (0, \frac{\pi}{2})$
Find (if exist) $\lim\limits_{R \rightarrow \infty}{|\cos z|}$
What happen if function $f(z)=|\cos z|$ when $\alpha=\frac{\pi}{2}$?
I have no idea how to do it. Any ideas? I will be so grateful.
Note that $$ \cos(x+iy)=\cos(x)\cosh(y)-i\sin(x)\sinh(y) $$ In your example, $Re^{i\alpha}=R\cos(\alpha)+iR\sin(\alpha)$, therefore, $$ \cos(Re^{i\alpha})=\cos(R\cos(\alpha))\cosh(R\sin(\alpha))-i\sin(R\cos(\alpha))\sinh(R\sin(\alpha)) $$ If $\sin(\alpha)\ne0$, as $R\to\infty$, $\cosh(R\sin(\alpha))\sim|\sinh(R\sin(\alpha))|\sim\frac12e^{R\,|\!\sin(\alpha)|}$.
Therefore, if $\sin(\alpha)\ne0$, as $R\to\infty$, $$ \begin{align} \left|\cos(Re^{i\alpha})\right| &\sim\frac12\left|e^{iR\cos(\alpha)}\right|e^{R\,|\!\sin(\alpha)|}\\ &=\frac12e^{R\,|\!\sin(\alpha)|}\\ \end{align} $$ and so $$ \lim_{R\to\infty}\left|\cos(Re^{i\alpha})\right|=\infty $$ However, if $\sin(\alpha)=0$, then as $R\to\infty$, $$ \left|\cos(Re^{i\alpha})\right|=|\cos(R)| $$ and so $$ \lim_{R\to\infty}\left|\cos(Re^{i\alpha})\right|\text{ does not exist} $$