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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?

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    $\begingroup$ what does a complex number tending to infinity mean??? i.e you cant compare reals over complex. I think you mean mod(z) tending to infinity $\endgroup$ – Prayas Agrawal Apr 1 '17 at 9:06
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    $\begingroup$ @PrayasAgrawal $\infty$ is a perfectly fine point in complex analysis.. $\endgroup$ – Stefano Apr 1 '17 at 9:08
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    $\begingroup$ Hmm..... Interesting $\endgroup$ – Prayas Agrawal Apr 1 '17 at 9:09
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    $\begingroup$ @Stefano I think he is possibly refering to the fact that the complex numbers aren't ordered, but in fact the notation is used to mean taking both the limit on the real axis and on the imaginary axis $\endgroup$ – ÍgjøgnumMeg Apr 1 '17 at 9:18
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    $\begingroup$ @Ragnar1204: Presumably, what you are trying to express is that the limit should be restricted to $\arg(z) = k$. $\endgroup$ – user14972 Apr 1 '17 at 11:17
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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).

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Hint

$$|e^z| = \left| e^{|z|(\cos k + i \sin k)}\right| = \left |e^{|z| \cos k} \right| \left| e^{i |z|\sin k}\right|.$$

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Hint:

write $e^z=e^{x+iy}=e^xe^{iy}$ so that : $|e^z|=e^x$

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