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Show analytically (finding the centre and radius) that $z(t)=\frac{1}{(1-i)^{-1}-t}=\frac{2}{1+i-2t}$ where $z(t)\in C $, that $z(t)$ traces out a circle in the complex plane as $t$ is varied.

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If you apply an inversion, the image of $\frac{2}{1 + i -2t}$ gets taken to the line $\frac{1+i}{2} - t$ which is the horizontal line passing through $\frac{i}{2}$. Undoing the inversion again, it's clear that the line gets taken to the circle with diameter $[-2i,\ 0]$.

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To find the radius, find the curvature of $\kappa(t)$ of $z(t)$ and shows that it's constant. The radius is then $r := \kappa(t)^{-1}$. To find the center $c$, find $\lambda \in \{-i,i\}$ such that $c=z(t)+r\lambda \frac{z'(t)}{||z'(t)||}$ is constant. Note that the second expression amounts to adding $r$ times the normal vector at $t$ to $z(t)$, which if $z(t)$ describes a circle should be the center.

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