Write . For real and , the denominator cannot vanish: if its zero is real, and if then . Taking the imaginary part after multiplying by givesThus . Alsoso it is conformal. Cross-multiplication shows that implies , proving injectivity.
The inverse Möbius transformation isIts coefficients are real and its determinant is again , so the same imaginary part identity shows that it maps into . Therefore every has a preimage in , proving surjectivity. Hence maps one-to-one and onto itself.
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