entire and injective function

Valentin Vornicu

Let $f$ be an entire function. Prove that if $f$ is injective then $f(z)=az+b$ , for some $a,b \in \mathbb{C}$ , $a\neq 0$ .

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luqash

if $f$ is injective, $f'$ never vanishes. By Liouville, $f'$ is constant. thus $f'(z) = a, f(z) = az + b$ .

**Posted:** Fri May 21, 2004 4:53 pm

Valentin Vornicu

hmm, it may be so, but I can think of an example of function that does not vanish at all, it is the derivative of an entire function, however, it is not constant. namely $f(z)=e^z$ . maybe you can explain some more your argument?

my soln. is the following: the Great Picard theorem tells us that if an entire function is not a polynomial, then it assumes each value (besides at most one) an infinite number of times.

since our $f$ is injective, it must be a polynomial of degree $k$ . But each polynomial of degree $k$ , assumes each complex value $k$ times exactly. So $k=1$ , and we are done.

**Posted:** Fri May 21, 2004 10:14 pm

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We all use math everyday: to forecast weather, to tell time, to handle money; we also use math to analyze crime, reveal patterns, predict behavior. Using numbers we can solve the biggest mysteries we know.

luqash

OK, i see my "argument" was a bit too short. Let my try another way: $g(z) = f(\frac 1z)$ is injective from $\mathbb C^* \cup {\infty}$ to $\mathbb C$ and holomorphic except at $0$ . It thus has a singularity at $0$ , which is removable or a pole of order at most $1$ by the injectivity of $g$ . $g$ 's Laurent expansion at $0$ therefore is $g(z) = zR(z) + b + \frac az$ , $R(z)$ being entire. Then $f(z) = g(\frac 1z) = \frac {R(\frac 1z)}z + b + az$ . Since $f$ is analytic at $0$ , $R(\infty) = 0$ . $R \equiv 0$ , because $R$ is entire and vanishes at infinity (Liouville finally!).

Ouf, I hope i got it right now.

**Posted:** Sat May 22, 2004 5:20 am

koo · P versus NP Offline Joined: 27 Dec 2004 Posts: 24

**Posted:** Thu Feb 10, 2005 5:40 am

ComplexZeta

An entire function that is not a polynomial has an essential singularity at $\infty$ , so consider $g(z)=f(1/z)$ and apply Picard's Great Theorem.

**Posted:** Thu Feb 10, 2005 4:00 pm

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