Proof that e^z is not a finite polynomial

[freefall111]

Homework Statement

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

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?

 

[micromass]

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

 

[capandbells]

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

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

 

[micromass]

Woow, I'm stupid today.

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

 

[freefall111]

e^(2pi*i) = cos(2pi) + isin(2pi) = 1.

e^z on the complex plane can never equal zero.

http://en.wikipedia.org/wiki/Exponential_function#Similar_properties_of_e_and_the_function_ez

 

[Ray Vickson]

Homework Statement

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

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?

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