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$f(z_1 z_2) = f(z_1) f(z_2)$ 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?

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    You can't delete a question as soon as there are answers. You can choose an answer to accept, if you wish to do so or leave the question as it is. That it is closed simply means that no more answers can be added.2011-07-18

3 Answers 3

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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 $b\in D$.

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    Thanks! I'm looking forward to the answers! I hope there will be an interesting answer (no matter if continuous or measurable is the hypothesis that ends up being used).2011-07-19
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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 $0$.

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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.