We were given a complex number problem to work on: if \[\mid s\mid =1\], then show that \[\Re(\frac{1}{s+1})=\frac{1}{2}\] I can prove this algebraically, but am having a little trouble with proving it geometrically. Can someone help me?

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TopNewestThe transformation \(s\mapsto \tfrac{1}{s+1}\) is a Mobius transformation, and Mobius transformations map circles/straight lines onto circles/straight lines. Thus the image of \(|s|=1\) is a circle/straight line \(L\) which passes through \(\infty\) (the image of the point \(-1\) on the original circle). Thus \(L\) is a straight line. Since the point \(2\) on the circle maps to the point \(\tfrac12\), \(L\) passes through \(\tfrac12\).

Mobius transformations also preserve the angles between lines. The circle \(|s|=1\) meets the real axis at right angles at \(1\). Hence \(L\) meets the real axis (the image of the real axis) at right angles at \(\tfrac12\). Thus \(L\) must be parallel to the imaginary axis, and we are done. – Mark Hennings · 3 years, 1 month ago

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– Edward Jiang · 3 years, 1 month ago

Thanks, appreciate it. :)Log in to reply

The set \(A = \{s+1 \ : \ |s| = 1\}\) is a circle with center \(1\) and radius \(1\). This circle passes through the origin. The set \(A' = \{\tfrac{1}{s+1} \ : \ |s| = 1\}\) is the inversion of \(A\). Under inversion, circles passing through the origin map to lines that do not pass through the origin. Now, you just need to show that the line is given by \(\mathfrak{R}(z) = \tfrac{1}{2}\). – Jimmy Kariznov · 3 years, 1 month ago

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