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Question Number 110504 by Rio Michael last updated on 29/Aug/20

Given a transformation , T: C → C ; z → ω   Show that if ω = ((z−i)/(z +1)) then z = ((ω + i)/(1−ω)). Hence  the image of the line ∣z−i∣ = ∣z + 2∣ under the transformation  T    the ω−plane is a circle with center (−2,−i) and radius (√(10)) .

$$\mathrm{Given}\:\mathrm{a}\:\mathrm{transformation}\:,\:\mathcal{T}:\:\mathbb{C}\:\rightarrow\:\mathbb{C}\:;\:{z}\:\rightarrow\:\omega \\ $$$$\:\mathrm{Show}\:\mathrm{that}\:\mathrm{if}\:\omega\:=\:\frac{{z}−{i}}{{z}\:+\mathrm{1}}\:\mathrm{then}\:{z}\:=\:\frac{\omega\:+\:{i}}{\mathrm{1}−\omega}.\:\mathrm{Hence} \\ $$$$\mathrm{the}\:\mathrm{image}\:\mathrm{of}\:\mathrm{the}\:\mathrm{line}\:\mid{z}−{i}\mid\:=\:\mid{z}\:+\:\mathrm{2}\mid\:\mathrm{under}\:\mathrm{the}\:\mathrm{transformation} \\ $$$$\mathcal{T}\:\:\:\:\mathrm{the}\:\omega−\mathrm{plane}\:\mathrm{is}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{center}\:\left(−\mathrm{2},−{i}\right)\:\mathrm{and}\:\mathrm{radius}\:\sqrt{\mathrm{10}}\:.\: \\ $$

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