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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?

vitamin d
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Ragnar1204
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    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 – oops Apr 01 '17 at 09:06
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    @PrayasAgrawal $\infty$ is a perfectly fine point in complex analysis.. – Stefano Apr 01 '17 at 09:08
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    Hmm..... Interesting – oops Apr 01 '17 at 09:09
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    @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 – Edward Evans Apr 01 '17 at 09:18
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    @Ragnar1204: Presumably, what you are trying to express is that the limit should be restricted to $\arg(z) = k$. –  Apr 01 '17 at 11:17
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    @ig: That's not true! Any neighborhood of $\infty$ contains points whose imaginary component is zero, and also points whose real component is zero. So to show a limit at $\infty$ exists, it is not sufficient to compute a limit as the magnitudes of the real and imaginary parts go to $+ \infty$ –  Apr 01 '17 at 11:23
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    @PrayasAgrawal: The sets with $|z| > M$ do indeed serve as neighborhoods of $\infty$. But it is also true that the single point $\infty$ of the projective complex numbers doesn't mesh especially well with the two separate points $\pm \infty$ of the extended real numbers. –  Apr 01 '17 at 11:28
  • Since they are talking about $\arg(z)=k$, I believe they are interested in $z=re^{ik}$. That is, $$\lim_{r\to\infty}\left|e^{re^{ik}}\right|$$ – robjohn Jul 13 '21 at 22:14

3 Answers3

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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).

E. Joseph
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Hint

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

Stefano
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Hint:

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

Emilio Novati
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