If $f$ is entire and for some $r>0$ we have $f(z)=Cz^n$ for a constant $C \in\mathbb C$ on $|z|=r$ then does it follow that $f(z)=Cz^n$ for all $z$? If $f$ is entire and for some $r>0$ we have $f(z)=Cz^n$ for a constant $C \in\mathbb C$ on $|z|=r$ then does it follow that $f(z)=Cz^n$ for all $z$? Since a circle is not open the identity principle doesn't apply so I'm not sure how to prove it, assuming it is even true. 
 A: An entire function has a Taylor expansion
$$f(z)=a_0+\sum\limits_{k=1}a_kz^k \tag{1}$$
where 
$$a_k=\frac{f^{(k)}(0)}{k!}=\frac{1}{2\pi i}\int\limits_{|z|=r}\frac{f(z)}{z^{k+1}}dz=\frac{1}{2\pi i}\int\limits_{|z|=r}\frac{Cz^n}{z^{k+1}}dz$$
We can see now that:


*

*if $\color{blue}{k<n} \Rightarrow$
$$\color{red}{a_k=}\frac{1}{2\pi i}\int\limits_{|z|=r}\frac{Cz^n}{z^{k+1}}dz=
\frac{C}{2\pi i}\int\limits_{|z|=r}z^{n-k-1}dz=
\frac{C}{2\pi i}\int\limits_{0}^{2\pi}(re^{it})^{n-k-1}rie^{it}dt=\\
\frac{Cr^{n-k}}{2\pi}\int\limits_{0}^{2\pi}e^{it(n-k)}dt=
\frac{Cr^{n-k}}{2\pi i}\cdot \frac{e^{it(n-k)}}{n-k}\bigg\rvert_{0}^{2\pi}=\color{red}{0}$$

*if $\color{blue}{k=n} \Rightarrow $
$$\color{red}{a_k=}\frac{1}{2\pi i}\int\limits_{|z|=r}\frac{Cz^n}{z^{n+1}}dz=
\frac{C}{2\pi i}\int\limits_{|z|=r}\frac{1}{z}dz=
\frac{C}{2\pi i}\int\limits_{0}^{2\pi}\frac{1}{re^{it}}rie^{it}dt=
\frac{C}{2\pi}\int\limits_{0}^{2\pi}dt=
\color{red}{C}$$

*if $\color{blue}{k>n} \Rightarrow $
$$\color{red}{a_k=}\frac{1}{2\pi i}\int\limits_{|z|=r}\frac{Cz^n}{z^{k+1}}dz=
\frac{C}{2\pi i}\int\limits_{|z|=r}\frac{1}{z^{k+1-n}}dz=
\frac{C}{2\pi i}\int\limits_{0}^{2\pi}\frac{1}{(re^{it})^{k+1-n}}rie^{it}dt=\\
\frac{C}{2\pi r^{k-n}}\int\limits_{0}^{2\pi}\frac{1}{e^{it(k-n)}}dt=
-\frac{C}{2\pi i r^{k-n}}\cdot \frac{1}{k-n}\cdot \frac{1}{e^{it(k-n)}}\bigg\rvert_{0}^{2\pi}=\color{red}{0}$$


and the result easily follows from $(1)$. 
A: You can use Cauchy formula to see that for $|z| < r$:
$$f(z)=\frac{1}{2 i \pi} \int_\mathcal{C} \frac{C \xi^n}{z-\xi}d \xi$$
where $\mathcal{C}$ is the circle or radius $r$.
So $f(z)=C z^n$ for all $|z| \leq r$ which is an open set.
