TITLE: Complex limit of an exponential.
QUESTION [5 upvotes]: For which values of $\arg(z)=k$ for $z\in\mathbb{C}$ does
$$\lim_{z\rightarrow \infty}|e^z|$$ exist? Consider $k$ constant.
I don't have any idea on how to do this. Anyone can please help me?

REPLY [6 votes]: You have 
$$\vert e^z\vert =\vert e^{x+iy}\vert=e^x$$
where $x$ is the real part of $z$ and $y$ is imaginary part.
So you have for all $k=\mathrm{arg}(z)\in(-\pi/2,\pi/2)$:
$$\lim_{\vert z\vert \to\infty} \vert e^z\vert =\lim_{x\to+\infty} e^x=+\infty$$
because $x\to\infty$ when $\vert z\vert\to\infty$ in that case (you assumed $k$ is constant in your question).
Otherwise, if $k=\pi/2$ or $k=-\pi/2$:
$$\lim_{\vert z\vert \to\infty} \vert e^z\vert =\lim_{x\to+\infty} 1=1.$$
And finally, if $k\in(\pi/2,2\pi/2)$:
$$\lim_{\vert z\vert \to\infty} \vert e^z\vert =\lim_{x\to+\infty} e^{-x}=0.$$
So the limit will always exist if you assume $k$ constant (it is different if you don't).