Let \(f: \mathbb{C} \rightarrow \mathbb{C}\) be an injective analytic (also called univalent) function. Show that there exist complex numbers \(a \neq 0\) and \(b\) such that \(f(z)=a z+b\).
Hint: Apply the Casorati-Weierstrass theorem to \(f(1/z)\).
Solution
Note that \(f\) is non-constant, since a constant function is extremely non-injective. Consider the singularity at \(\infty\):
If it is removable, then \(f\) is bounded outside of a large disc, and bounded inside of it as a continuous function on a compact set, making \(f\) entire and bounded and thus constant by Liouville.
If it is essential, then by Casorati-Weierstrass there is a large disc of radius \(R\) such that \(f(\bar{\DD_R}^c) \subseteq \CC\) is dense. By the open mapping theorem, \(f(\DD_R) \subseteq \CC\) is open, so by density it intersects \(f(\bar{\DD_R}^c)\), but \(\DD_R \intersect \bar{\DD_R}^c\) is empty so this contradicts injectivity.
So we can conclude \(\infty\) is a pole of some order \(N\), so \(f\qty{1\over z} = \sum_{0\leq k\leq N} c_k z^{-k}\) and thus \(f(z) = \sum_{0\leq k\leq N} c_k z^k\) is a polynomial of degree \(N\). However, a polynomial of degree \(N\) is generically \(N\)-to-one locally, so injectivity forces \(N=1\) and \(f(z) = c_0 + c_1 z\), where \(c_1\neq 0\) since \(f\) is nonconstant.
Solution: older
Write \(g(z) \da f(1/z)\), which has a singularity at \(z=0\). The claim is that this is a pole.
If \(z=0\) is a removable singularity, \(g\) is bounded on some closed disc \(\abs{z} \leq \eps\), so \(f\) is bounded on \(\abs{z} > \eps\). Moreover \(f\) is continuous and \(\abs{z}\leq \eps\), \(f\) is bounded on this disc. This makes \(f\) an entire bounded function and thus constant by Liouville, contradicting injectivity.
If \(z=0\) is essential, then by Casorati-Weierstrass pick a punctured disc \(D = \ts{\abs{z} \leq \eps}\) where \(g(D)\) is dense in \(\CC\). Writing \(D^c \da \ts{\abs{z} > \eps}\), this means that \(f(D^c)\) is dense. But \(U\da \ts{\abs{z} < \eps}\) is open and by the open mapping theorem \(f(U)\) is open, so by density there is a point \(w\in f(D^c) \intersect f(U)\) while \(U \intersect D^c = \emptyset\), again contradicting injectivity.
So \(z=0\) is a pole of \(g\), and \(g\) admits a Laurent expansion \[ g(z) = \sum_{k\geq -N} c_k z^k .\] Since \(f\) is entire, it equals its Laurent expansion at \(z=0\), so equating the two series yields \[ f(z) = \sum_{k\geq 0} d_k z^k &\implies g(z) = \sum_{k\geq 0} {d_k \over z^k} = \sum_{1\leq k\leq N} {c_k\over z^k} + \sum_{k\geq 0} c_k z^k \\ &\implies \sum_{k\geq 0} c_k z^k = 0 \\ &\implies f(z) = \sum_{0\leq k \leq N} c_k z^k ,\] making \(f\) a polynomial of degree at most \(N\).
Now \(f\) can not be degree zero, since constant maps are not injective. Moreover \(f\) can not be degree \(N\geq 2\), since any polynomial of degree \(N\) has \(N\) roots in \(\CC\) by the fundamental theorem of algebra, and any two distinct roots will be points where injectivity fails. Finally, ruling out the case of roots with multiplicity, if \(f(z) = c(z-a)^N\), then \(f\) has exactly \(N\) preimages in a neighborhood of \(a\). Letting \(p\) be any such point, we can find \(N\) complex points mapping to it: \[ p = c(z-a)^N &\implies {p\over c} = (z-a)^N \\ &\implies \qty{p\over c}^{1\over N}\zeta_N^k = z-a \quad k=0,1,\cdots, n-1 \\ &\implies z_k\da \qty{p\over c}^{1\over N}\zeta_N^k + a \mapsvia{f} p .\]
So \(f\) must be degree exactly 1, i.e. \(f(z) = az+b\).