Originally Posted by

**shawsend** Ok, I'd like to try:

Prove that if $\displaystyle f(z)=0$ on a subset of the real axis $\displaystyle S=\{z: a<z<b\}$, then $\displaystyle f(z)=0$ throughout $\displaystyle \mathbb{C}$.

Since $\displaystyle f(z)$ is analytic, then we can write for some $\displaystyle c\in S$:

$\displaystyle \begin{aligned}f(z)&=\sum_{n=0}^{\infty} a_n(z-c)^n \\ &=a_m(z-c)^m+a_{m+1}(z-c)^{m+1}+\cdots \\

&=(z-c)^m\left[a_m+(z-c)Q(z)\right]

\end{aligned}$

where $\displaystyle Q(z)$ is analytic and bounded in some neighborhood of $\displaystyle c$, that is $\displaystyle |Q|<M$ for $\displaystyle |z-c|<R$, and $\displaystyle m$ is the first non-zero term. Now:

$\displaystyle |f(z)|=|z-c|^m\left|a_m+(z-c)Q(z)\right|$

so that:

$\displaystyle |f(z)\geq |z-c|^m\left||a_m|-|z-c||Q|\right|$

Now recall that we have a dense set of points over which $\displaystyle f(z)=0$. Therefore, we can find a $\displaystyle z_i\in S$ such that $\displaystyle |z_i-c|<\frac{|a_m|}{M}$. This means:

$\displaystyle |z_i-c||Q|<\frac{|a_m|}{M} M<|a_m|$

This would imply $\displaystyle |f(z_i)||\geq |z_i-c||a_m|>0$ since $\displaystyle z_i\ne c$ and $\displaystyle a_m\ne 0 $ but by assumption $\displaystyle f(z_i)=0$. This must mean we've reached a contradiction and all $\displaystyle a_m=0$ so that $\displaystyle f(z)=0$.

I'll leave it to you to show the function $\displaystyle f-g=0$ on the same domain implies $\displaystyle f=g$ since $\displaystyle h(z)=f(z)-g(z)$ can still be written as $\displaystyle \sum_{n=0}^{\infty} a_n(z-c)^n$ for some (new) set of $\displaystyle a_n$.

This may need a little more work . . .