Multiplicative Analytic Functions? 
Possible Duplicate:
$f(z_1 z_2) = f(z_1) f(z_2)$ for $z_1,z_2\in \mathbb{C}$ then $f(z) = z^k$ for some $k$ 

How can we characterize the analytic functions defined in the open unit disc $D\subset\mathbb{C}$ that satisfy $f(ab)=f(a)f(b)\text{ }$  for all $a,b\in D$.
What happens if we consider larger domains?
 A: Differentiating with respect to $a$ yields $f^{(n)}(ab)b^n=f^{(n)}(a)f(b)$ for all $a$ and $b$ in $D$ and for each positive integer $n$.  In particular, $f^{(n)}(0)b^n=f^{(n)}(0)f(b)$ for all $b$ in $D$ and $n\in\mathbb N$.  If $f^{(n)}(0)=0$ for all $n>0$, then $f$ is constant, $0$ or $1$.  If $f^{(n)}(0)\neq 0$ for some $n>0$, then $f(b)=b^n$ for all $b\in D$.
A: We have $f(0) = f(0)f(b)$ hence, if $f$ is not the constant function equal to $1$, we have $f(0) =0$. If $f$ is not the constant function equal to $0$, then we can find $k\in\mathbb{N}^*$ such that $f(z) = z^kg(z)$ with $g(0)\neq 0$ and $g$ analytic. We get for $a,b\neq 0$ that $g(a)g(b) =g(ab)$ and by continuity for all $a$ and $b$. We have $g(0)=g(0)g(b)$ hence $g(z)=1$ for all $z$. Finally the only solutions are $f(z)=0$, $f(z)=1$ and $f(z)=z^k, k\in\mathbb{N}$. 
We only used the fact that the unit disc is connected; the result can be extended to each open connected subset of $\mathbb C$ which contains $0$.
A: Suppose $f(0)\neq 0$.  Then since $f(0)=f(0)f(b)$ for any $b$, we see that $f(b)=1$ for every $b$, so $f$ is identically the constant function.  Suppose $f(0)=0$.  Then $f(z)=z^m g(z)$ for some $g$ with $g(0)\neq 0$.  Since $f$ and $z^m$ are multiplicative, so is $g(z)$.  But by the first part, $g(z)=1$ for all $z$ and is identically the constant function.  Hence $f(z)=z^m$.
Conclusion:  $f(z)=z^m$ for some nonegative integer.
