# Homework Help: Proof that e^z is not a finite polynomial

1. Mar 31, 2012

### freefall111

1. The problem statement, all variables and given/known data

Prove that the analytic function e^z is not a polynomial (of finite degree) in the complex variable z.

3. The attempt at a solution

The gist of what I have so far is suppose it was a finite polynomial then by the fundamental theorem of algebra it must have at least one or more roots. e^z can never equal zero for hence this is a contradiction.

Is it okay for me to apply the fundamental theorem of algebra like this or am I kind of using a bit too much machinery here?

2. Mar 31, 2012

### micromass

Your proof is wrong. Indeed: $e^{2\pi i}=0$, so the function DOES have a root!!

3. Mar 31, 2012

### capandbells

Wait, what? $e^{i2\pi} = 1$

4. Mar 31, 2012

### micromass

Woow, I'm stupid today.

I'm sorry, your proof is alright!! I obviously need to get some sleep.

5. Mar 31, 2012

### freefall111

6. Mar 31, 2012

### Ray Vickson

That looks like too much machinery (at least for my tastes); I would rather just argue that a polynomial of degree n has (n+1)st derivative = 0 identically, and ask whether that can happen for exp(z).

RGV