Question #0f1ea

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Question #0f1ea

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Answer 1 · 2016-12-29T14:22:44

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#

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