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I have come across a question in a textbook "Sketch the region in an Argand diagram where $\lim_{z\to \infty} |e^{z^3}|=0$

The solution in the book begins "This will only be satisfied if $\mathrm{Re}(z^3)$ is negative", without any justification as if it is obvious, but I cannot see why this is the correct condition.

Can anyone explain please? Thank you!

Noa Even
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Tharaib
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2 Answers2

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If $z=x+iy$ with $x,y\in\mathbb R$, then

$$|\exp(z)| = |\exp(x+iy)|=|\exp(x)\exp(iy)| = \exp(x)\,.$$

Astyx
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  • Thank you for the very fast answer! But I'm afraid I still don't see how this leads to it being the correct condition for the required behaviour for $|exp(z^3)|$. I'm just failing to see the line of reasoning from there to the condition in the question. Sorry – Tharaib Jan 03 '22 at 09:07
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    $\exp(x)$ is the exponential of a real number. If $x\to -\infty$ it goes to $0$, if $x\to \infty$ it goes to $\infty$ – Astyx Jan 03 '22 at 09:13
  • ‍♂️ Ugh I guess I couldn't see the wood for the trees. Thank you for taking the time to explain something so simple to someone who was being very stupid! – Tharaib Jan 07 '22 at 01:59
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If we denote $z^3=a+ib$, where $a=Re(z^3), b=Im(z^3)$, then $|e^{a+ib}|=|e^ae^{ib}|=|e^a|$. In our case $|e^{a+ib}|=|e^a|\rightarrow 0$ and $e^a$ is a real exponential, so $|e^a|=e^a\rightarrow 0$ for $|a+ib|\rightarrow\infty$ iff $a\rightarrow -\infty$. So we have the necessary and sufficient condition $Re(z^3)\rightarrow-\infty$.

FreeFunctor
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  • I just want to add that $|z|\to \infty$ does not imply $Re(z)\to \pm \infty$ or $Im(z)\to \pm \infty$, take $z=t\exp(it)$ with $t\in\mathbb R$ for a counterexample. – Astyx Jan 03 '22 at 09:18
  • Oh right! I delete the useless row. – FreeFunctor Jan 03 '22 at 09:19