About the behavior of $e^z = e^x e^{iy}$ as
(a) $x$ tends to $- \infty$;
(b) $y$ tends to $ \infty$.
For $y$ tends to $ \infty$ I got that $e^z$ just rotates around the circle with radius $e^x$ over and over again.
About the behavior of $e^z = e^x e^{iy}$
2
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real-analysis
complex-analysis
exponential-function
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0Yes, $e^z$ is periodic in the imaginary axis direction. The factor $e^x$ gives a modulus (absolute value) which is increasing with $x$. – 2017-01-28
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0The limit when $y\to\infty$ is undefined. A more interesting question is if $y\to\infty$ in a very specific way, following a specific sequence. Then the result is not necessarily undefined. – 2017-01-28
2 Answers
1
The exponential map send lines parallel to $x-$axis (so $y$ is constant) into half lines starting from the origin, in particular you can verify that when $x\rightarrow -\infty$ and $y$ remain constant, then $e^z$ approach to zero along that half line. Infact $e^z=e^{x+iy}=e^xe^{iy}$, if $y$ is constant, then the argument remain constant, and the modulus decrease if $x$ goes to $-\infty$.
1
for $e^z=re^{iy}$, When $x\to-\infty$ the value of $r=e^{x}\to0$ that is with every constant argument, modulus $r$ will be smaller and smaller.