If $z,s\in \mathbb{C}$ ( with norm less than $1$) then show that $\left|\frac{1-s/z}{1/z-\overline{s}}\right|<1$. I'm hoping for an algebraic proof but possibly also a proof by geometric means if possible ( even with a picture if possible).
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1An analytic way: Have a look at the automorphism group of the unit disk, and use the Schwarz lemma. – Jan 18 '19 at 01:22
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I don't see how to apply that theorem since $|f(0)|=|s|$ which isn't necessarily zero. – Squirtle Jan 18 '19 at 01:37
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You don't apply it to this function directly, but rather start by recognizing this as the value of an automorphism of $\mathbb{D}$ evaluated at a point. – Jan 18 '19 at 02:11
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2Possible duplicate of Show that $\left|\frac{\alpha - \beta}{1-\bar{\alpha}\beta}\right| < 1$ when $|\alpha|,|\beta| < 1$ – Martin R Jan 18 '19 at 06:16
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Also: https://math.stackexchange.com/q/1630930/42969, https://math.stackexchange.com/q/342181/42969. – Martin R Jan 18 '19 at 06:17
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Your inequality is equivalent to $|1 - s/z| < |1/z - \bar{s}| \iff |z - s|^2 < |1 - \bar{s} z|^2$. Expanding yields $$ z \bar{z} + s \bar{s} - z \bar{s} - \bar{z} s < 1 - z \bar{z} s \bar{s} - z \bar{s} - s \bar{z}$$ So we need to show $|z|^2 + |s|^2 < 1 + |s|^2|z|^2$. But this is equivalent to $0 < (1 - |z|^2) ( 1 - |s|^2)$ which is clearly true since $|z|, |s| < 1$.
Martin R
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Carlos Esparza
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