Question

Question #0f1ea

Answer 1

See below.

Explanation:

Calling `z=x+iy` we have `abs(z)=sqrt(z bar(z)) = sqrt((x+iy)(x-iy))=sqrt(x^2+y^2)` so
`abs(z)=1` corresponds to a circle centered at the origin of coordinates, with radius 1. Squaring both sides

`x^2+y^2= 1`