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How to prove that every bijective rational function (over the field of complex numbers) is a linear fractional transformation ?


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up vote 5 down vote accepted

Say $f(x)/g(x)=t$ for a given $t$ then consider the polynomial $h(x)=f(x)-tg(x)$. Over the complex numbers $h$ will more than one root if $f, g$ are not of degree 1. So the only bijections are the linear fractional transformations.

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I love this answer! It is flawless. – MJD Sep 19 '12 at 13:59

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