Is there a holomorphic function on the unit disk that satisfies $f(\frac{1}{n})=\frac{1}{n^{3/2}}$? Is there a holomorphic function on the unit disk that satisfies $f(\frac{1}{n})=\frac{1}{n^{3/2}}$?
I know that I should use identity theorem. How do I apply it here?
 A: Assume that $f$ is holomorphic with the indicated property; we'll derive a contradiction.
Since $f$ is continuous, $f(0)=0.$ It follows that $f(z)^2=z^3$ for all $z$ in the set $\{\frac1{n}\mid n\in \mathbb{Z}^+\}\cup \{0\}.$ This set has an accumulation point, so, by analytic continuation, $f(z)^2=z^3$ for all $z$ in the unit disk.
Fix any value $\alpha$ such that $0\lt\alpha\lt 1,$ and define a function $g\colon\mathbb{R}\to\mathbb{C}$ by setting $g(\theta)=e^{-3i\theta/2}\,f(\alpha^2 e^{i\theta}).$
We have $g(\theta)^2 = e^{-3i\theta}\,\alpha^6 e^{3i\theta}=\alpha^6,$ so $g(\theta)=\pm \alpha^3$ for all $\theta\in\mathbb{R}.$
The function $g$ is continuous, and the range of a continuous function on a connected set must be connected, so either $g(\theta)=\alpha^3$ for all $\theta\in\mathbb{R}$ or $g(\theta)=-\alpha^3$ for all $\theta\in\mathbb{R}.$
But $g(0)=f(\alpha^2)$ and $g(2\pi)= e^{-3i\pi}f(\alpha^2 e^{2\pi i})=-f(\alpha^2),$ which gives us our contradiction. (Note that $f(\alpha^2)$ can't equal $0,$ because of the identity $f(z)^2=z^3.)$

Edit comment: The original proof assumed that we were working in a domain containing the closed unit disk, since we used the value of $f(e^{i\theta}).$ This revision corrects this by introducing the multiplier $\alpha$; we now only need that $f$ is holomorphic on the (open) unit disk, as the problem actually stated.
A: it agrees with $$g(z) = z^{3/2}$$ on the set $1/n.$ 
$g$ cannot be made holomorphic at zero but $f$ must agree with it. So the answer is no.
it's not as simple as applying the identity theorem since $g$ is not holomorpic at zero. 
However, note that $f'(0)= f(0)=0$ if $f$ is holomorphic by considering limit of finite differences.
 So $f$'s power expansion has first two terms zero so
$|f(z)| \leq c|z|^2$ near $0$ but this is not compatible with the stated property of $f.$
So $f$ cannot be holomorphic.
