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Same as my previous question except domain is complex. I tried assuming that the function was analytic, so for $z_1=z_2=z$ , $f(z^2) = f(z)^2$ $$\sum_{n=0}^\infty a_n z^{2n}=\left(\sum_{n=0}^\infty a_n z^n\right)^2$$ and try to solve it. Ideally, I should have something like $a_n=0$ for all but one k. and that should prove it.

This question, with $z_1=z_2$ was asked today but I have not been able to understand their answers.

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So, the requirement that $f$ be analytic is necessary, since $f(z)=\bar{z}$ is an example which is not analytic. – Thomas Andrews Jun 9 '11 at 19:05
Also, you don't want $z_n$ in your expansion, just $z^n$: $\sum_{n=0}^\infty a_n z^{2n}=\left(\sum_{n=0}^\infty a_n z^n\right)^2$ – Thomas Andrews Jun 9 '11 at 19:06
@Thomas I fixed that typo, thanks. – kuch nahi Jun 10 '11 at 2:19
up vote 9 down vote accepted

This question is very similar to the other.

Notice that the constant zero function is a solution. For the remainder of this answer, I assume that $f$ is not identically $0$.

First, let $z_1=0$. Then $f(0)=f(0)f(z_2)$ for all $z_2$. Lets split into cases since either $f(0)=0$ or $f(0)\neq 0$.

Case 1: Suppose $f(0)\neq 0$. Then in the equation $f(0)=f(0)f(z_2)$ we can divide through by $f(0)$ and conclude that $f(z_2)=1$ for all $z_2$, so $f$ must be the constant function $1$.

Case 2: Suppose $f(0)=0$. As $f$ is analytic, we may write $f(z)=z^k g(z)$ for some analytic $g$ where $g(0)\neq 0$. This function $g$ must satisfy the same functional equation as $f$ (why?). Hence by case 1, it follows that $g=1$, and we conclude $f(z)=z^k$.

Hope that helps,

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Let $k=f'(1)$. Define $f_a(z)=f(az)=f(a)f(z)$. Differentiating by $z$ we get:

$$f_a'(z) = a f'(az) = f(a) f'(z)$$

Setting $z=1$, we get that $a f'(a) = k f(a)$ for arbitrary $a$. Writing this as an equality of power series and we get, that $k$ must be a non-negative integer, and that $f(a)=a^k$, or that $f(a)=0$ for all $a$.

[This assumes that $f$ is defined on $1$ and $0$. Obviously, there are examples where it isn't defined on $z=0$ - the cases where $k<0.$]

If $f(1)$ is defined, then we don't need $f(0)$ because we can rewrite the above as: $f'(a)/f(a) = k/a$ when $a\neq 0$ and $f(a)\neq 0$. But the left side is the derivative of $\log f(z)$ for some branch of the natural log, and the right side is the derivative of $k \log z$ for some branch of $\log$. So $f(z) = e^{\log f(z)} = K e^{k \log z} = K z^k$ in some region around $a$, and hence all of the domain of $f$ (assuming the domain is connected?)

If $f(a)=0$, on the other hand, then $f(ax)=0$ for all $x$ near $1$, which would mean that $f$ is zero on some ball around $a$, so $f$ is identically zero.

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