# 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