$f:\mathbb{C}\rightarrow\mathbb{C}$ entire function and $f(z)=u(x)+iv(y)$ then $f$ is a polynomial

Question

I was going through my introduction to complex analysis homework, when I came across this exercise:

If $f:\mathbb{C} \rightarrow \mathbb{C}$ is an entire function of the form $f(z)=u(x)+iv(y)$, prove that $f$ is a polynomial.

I've got completely stuck on this one. I think it might be something to do with $f$ being analytical, but I'm not so sure, because I was sick and couldn't watch the class. ):

Any hints are appreciated, thanks.

score 10 · Accepted Answer · answered Sep 14 '11 at 01:58

10

Hint : Write down the Cauchy-Riemann equations and see what pops out.

answered Sep 14 '11 at 01:58

Adam Smith

4,320

I had written them before and didn't see anything that would imply $f$ being a polynomial.
I have that $\frac{du}{dx}=\frac{dv}{dy}$ and that the other partials are $0$.

I must be missing something obvious. ):
– Leonardo Fontoura Sep 14 '11 at 02:08
You should think about what it means for a partial to be $0$. Also, remember that $u$ only depends on $x$ and $v$ only depends on $y$. – Adam Smith Sep 14 '11 at 02:15
I had completely overlooked that, thanks. I'll try again. – Leonardo Fontoura Sep 14 '11 at 02:16
Now it became completely obvious! Thank you! (: – Leonardo Fontoura Sep 14 '11 at 02:17

$f:\mathbb{C}\rightarrow\mathbb{C}$ entire function and $f(z)=u(x)+iv(y)$ then $f$ is a polynomial

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