Sketch the set of complex numbers on the complex plane.

Question

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?

Answer 1

$e^z = e^{x+iy} = e^xe^{iy}\\ |e^x| = e^x\\ |e^{iy}| = 1$

Answer 2

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).