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

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

Commented by MJS last updated on 13/Dec/19

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

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