# Sketch the set of complex numbers on the complex plane.

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?

• What makes you think of this particular circle ?
– user65203
Jan 16 '20 at 7:53
• 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. Jan 16 '20 at 7:57
• Well, do not believe that the answer is always a circle. And in any case, I don't see a justification for the center $(1,0)$.
– user65203
Jan 16 '20 at 7:58
• $|z-z_0|<r$ 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. Jan 16 '20 at 8:00

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