TITLE: Recognize a circle in the complex plane
QUESTION [0 upvotes]: I have this equation: $z \in \mathbb{C}; \frac{1}{z} - \frac{1}{\bar{z}} = -i$
and transformed it into $2b = a^2 + b^2$ with $a=Re(z)$ and $b=Im(z)$
The original equation has a circle around $(0,1)$ with radius 1 as solution set. How can I read this from my equation?

REPLY [0 votes]: Cirrcle whose center is at $(-g,-f)$ with radius $\sqrt{f^2+g^2-c}$ can be expressed as
$$z\bar z+z_0\bar z+\bar z_0 z+c=0$$ is one form of equation of circle, whose center is at $z=-z_0=-g-if$ and radius is $r=\sqrt{z_0 \bar z_0 -c}$ In rhe given question
$$\frac{1}{z}-\frac{1}{\bar z}=-i \implies z\bar z-i\bar z+iz=0 \implies z_0=-i$$
So the center of this circle is at $z=-z_0=i=0+i$ or $(0,1)$, its radius is $\sqrt{-i.i}=1$
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