Graph complex inequality $|1/z|<1$, when $z \in \mathbb{C}$

Question

Problem

Graph complex inequality $|1/z|<1$, when $z \in \mathbb{C}$

Attempt to solve

if i set $z=x+iy$ when $(x,y) \in \mathbb{R}, z \in \mathbb{C}$

$$ |\frac{1}{x+iy}|<1 $$

Trying to multiply denominator and nominator with complex conjugate.

$$ |\frac{x-iy}{(x+iy)(x-iy)}| <3 $$

$$ |\frac{x-iy}{x\cdot x + x(-iy) + iy \cdot x + iy (-iy)}| <3$$

$$ |\frac{x-iy}{x^2-xyi+xyi-(iy)^2}|<3 $$

Utilize the fact that $i^2=-1$

$$ |\frac{x-iy}{x^2+y^2}|<3 $$

$$ |\frac{x}{x^2+y^2}-\frac{iy}{x^2+y^2}|<3 $$ $$ \sqrt{(\frac{x}{x^2+y^2})^2+(-\frac{iy}{x^2+y^2})^2}<3 $$ $$ (\frac{x}{x^2+y^2})^2+(-\frac{iy}{x^2+y^2})^2<3^2 $$

$$ \frac{x^2}{x^4+2x^2y^2+y^4}+\frac{i^2y^2}{x^4+2x^2y^2+y^4}<3^2 $$

$$ $$

I tried to plot the inequality with wolfram alpha and i got area outside of circle in $(0,0)

Not quite sure if this is correct.

Answer 1

Why not just write $$ |1/z| < 1 \quad\iff\quad |z| > 1 \quad\iff\quad x^2+y^2 > 1. $$

Answer 2

|z| > 1 that means outside the circle of radius 1 , centered at (0,0)