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Question: Sketch the set of complex numbers, z, that satisfy the modulus of $e^z<\frac{1}{2}$.

Answer: would I just plug-in 0 to find that the center of the circle is at (1,0) and has a radius of $\frac12$? I feel like that is too simple, can someone confirm or help me, please?

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  • $\begingroup$ What makes you think of this particular circle ? $\endgroup$
    – user65203
    Jan 16 '20 at 7:53
  • $\begingroup$ This is just my thought on this problem based on the example problems we solved in class. For example: $|(z+1+2i)|<2$ is a circle with a center at (-1,-2) and a radius of 2. $\endgroup$ Jan 16 '20 at 7:57
  • $\begingroup$ Well, do not believe that the answer is always a circle. And in any case, I don't see a justification for the center $(1,0)$. $\endgroup$
    – user65203
    Jan 16 '20 at 7:58
  • $\begingroup$ $|z-z_0|<r$ define the disc (and not the circle) centered at $z_0$ and radius $r$. But here you have $|e^z-0| < \frac 1 2$, then it is different. $\endgroup$
    – nicomezi
    Jan 16 '20 at 8:00
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$e^z = e^{x+iy} = e^xe^{iy}\\ |e^x| = e^x\\ |e^{iy}| = 1$

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The required region is the open half-plane $x<\ln \frac{1}{2}$. Note that $\ln\frac{1}{2}\approx-0. 693$.

The reason is that $|e^{x+iy}|=|e^x|$ and $e^{\ln\frac{1}{2}}=\frac{1}{2}$ (and $e^x$ is a strictly increasing function).

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