Suppose $f$ is entire and one-to-one. Show that $f(z)=az+b$. [duplicate]

My instinct is that it may be proved by Liouville's extended theorem. But how to do so? Or are there any other methods?

Thanks!

marked as duplicate by Julien, Ayman Hourieh, lhf, Potato, Ittay WeissMay 27 '13 at 0:34

• Consider the image of the unit circle. It has a winding number of 1. – Calvin Lin May 26 '13 at 23:59
• If $f(z)$ is not $b + az$ then $f(z) = b + az + cz^2 + \cdots$. And $z\mapsto z^2$ is not one-to-one. – Michael Hardy May 27 '13 at 0:02
• @MichaelHardy I don't see how your hint leads to an easy solution. Could you expand? – Julien May 27 '13 at 0:11
• @CalvinLin I don't see how that leads to a solution. Could you explain? – Potato May 27 '13 at 0:13
• @Potato That's what I call team work. – Julien May 27 '13 at 0:14

Hint: Consider $f(\frac{1}{z})$ and examine the behavior of the singularity at $0$. It can't be an essential singularity (see this using the Weierstrass-Casorati theorem on the behavior near an essential singularity along with the open mapping theorem). So it is either removable or a pole. Then $f(\frac{1}{z})$ has a finite number of negative powers of $z^{-1}$ in its Laurent expansion, so $f(z)$ is a finite power series and must be a polynomial. If the polynomial has degree greater than $1$, then it has $2$ or more roots, contradicting the one-to-one hypothesis.
Note that this gives you the automorphism group of $\mathbb C$.
• @LebronJames If it has a root with multiplicity $2$, then locally at that root it looks like $w=z^2$, which is clearly not injective. Ahlfors provides a rigorous explanation of this somewhere his book. Essentially, WLOG we may assume the root is at zero, so $f(z)=z^2(1+\dots)$, and then you can take a square root locally since the second factor is nonzero around $0$. – Potato Dec 18 '14 at 6:47