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Question Number 75580 by lalitchand last updated on 13/Dec/19

Define the inversion transformation of Z=a + ib where i is imaginary line passing through the circle of inversion.

$$\mathrm{Define}\:\mathrm{the}\:\mathrm{inversion}\:\mathrm{transformation} \\ $$$$\mathrm{of}\:\mathrm{Z}=\mathrm{a}\:+\:\mathrm{ib}\:\mathrm{where}\:\mathrm{i}\:\mathrm{is}\:\mathrm{imaginary}\:\mathrm{line}\: \\ $$$$\mathrm{passing}\:\mathrm{through}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{inversion}. \\ $$

Commented by MJS last updated on 13/Dec/19

it′s not clear. i is an imaginary line? I guess you mean the inversion at the circle of unity? in this case Z^∗ =(a/(a^2 +b^2 ))+(b/(a^2 +b^2 ))i or if Z=re^(iθ) ⇒ Z^∗ =(1/r)e^(iθ)

$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{clear}.\:\mathrm{i}\:\mathrm{is}\:\mathrm{an}\:\mathrm{imaginary}\:\mathrm{line}? \\ $$$$\mathrm{I}\:\mathrm{guess}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{the}\:\mathrm{inversion}\:\mathrm{at}\:\mathrm{the}\:\mathrm{circle} \\ $$$$\mathrm{of}\:\mathrm{unity}? \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case} \\ $$$${Z}^{\ast} =\frac{{a}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\frac{{b}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\mathrm{i} \\ $$$$\mathrm{or}\:\mathrm{if}\:{Z}={r}\mathrm{e}^{\mathrm{i}\theta} \:\Rightarrow\:{Z}^{\ast} =\frac{\mathrm{1}}{{r}}\mathrm{e}^{\mathrm{i}\theta} \\ $$

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