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I'm still new to complex functions and not very confident yet, so I wonder if you all can check if I've understood this problem correctly:

Suppose we take the limit of of $e^z$ as $|z| \to \infty$ along a half line $L$ (which I assume is a line from the origin to infinity). For which $L$ does this limit exist (can be finite or infinite)?

So, let $z=x+iy$. Then we can parametrize an arbitrary $L$ as those $z$ such that $x=r\cos{\theta_o}$ and $y=r\sin{\theta_o}$ for $r\in[0,\infty)$ for some fixed $\theta_o \in [0,2\pi)$. Then $|z| \to \infty$ as $r \to \infty$.

We see that $e^z = e^x (\cos{y} + i \sin{y})=e^{r\cos{\theta_o}}(\cos(r\sin{\theta_o})+i\sin(r\sin{\theta_o}))$

If $\theta_o = k\pi +\frac{\pi}{2}$ where $k \in \mathbb{Z}$, then as $r \to \infty$ the value of $e^z$ rotates around the unit circle and has no limit, while if $\theta_o \in (\frac{\pi}{2},\frac{3\pi}{2})$ then $|e^z|=|e^{r\cos{\theta_o}}|$ which goes to $0$ because $\cos{\theta_o}$ is negative. **

Otherwise, on the far right hand side of this equation, the complex number in the parenthesis, $(\cos(r\sin{\theta_o})+i\sin(r\sin{\theta_o}))$, has modulus $1$ and the argument is periodic in $r$. The complex number to the left of the parenthesis, $e^{r\cos{\theta_o}}$, is positive and real valued. Thus, the product has modulus equal to $e^{r\cos{\theta_o}}$, which goes to infinity because $\cos{\theta_o}>0$, and the argument never settles on a value. In this case, we still say that the limit is infinity, correct?

So, to summarize, for $L$ characterized by $\theta_{o} \in (\frac{\pi}{2},\frac{3\pi}{2})$, $e^z \to 0$; if $\theta_o = k\pi +\frac{\pi}{2}$ where $k \in \mathbb{Z}$, then the limit does not exist; and otherwise, $e^z \to \infty$ along $L$.

** Edited to reflect comments.

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Reconsider what happens if $\theta_0=\pi$. – Julián Aguirre Jan 24 '12 at 18:34
Thanks for pointing that out. – user18297 Jan 24 '12 at 19:36
Correct, that the modulus is $\exp(r\cos(\theta_0))$. But the limit $r\to\infty$ exists for some more $\theta_0\dots$ – Dirk Jan 24 '12 at 19:47

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