<s>[INST] Graph complex inequality $|1/z|<1$, when $z \in \mathbb{C}$ 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.
 [/INST] |z| > 1 
that means outside the circle of radius 1 , centered at (0,0)
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