Sketching the magnitude of a complex number

Question

I have the given inequality $|z+1|<2$ and I have to sketch this as a set. I take

$|x+iy+1|=\sqrt{x^2+y^2+1^2}$

$\sqrt{x^2+y^2+1}<2$

$x^2+y^2<3$

Sketching this should be a circle with radius $\sqrt{3}$ in the complex plane, but according to wolframalpha the right sketch is something very different:

What is correct here, and what have I misunderstood?

Answer 1

You calculated the magnitude incorrectly. Note that $1$ is real, so part of the real part of $z + 1$. Therefore, you get with $z = x+ iy$

$$| z + 1| = \sqrt{(x+1)^2 + y^2}.$$

Hence, all the points satisfying the inequality equivalently satisfy

$$(x+1)^2 + y^2 < 4.$$

This gives you the interior of a circle with midpoint $-1$ and radius $2$ on the complex plane.

As said, the WolframAlpha plot, interprets $z$ as being a real variable.

Still, you can see that this (on the real line) coincides with the plot (i.e. that all real solutions are in $(-3, 1)$).

Answer 2

The set $\{z \in \mathbb{C}: |z+1| <2\}$ is the set of complex number whose distance to -1 is less than 2, hence an open circle centered at $z_0=-1$ and radius 2.