Does there exist an analytic function $f(\frac{1}{n})^2 = \frac{n}{n+1}$ on (a) the complex plane and (b) the unit circle? If yes, how many? If no, why not?
I know I can write the function as: $f(\frac{1}{n}) = \pm \sqrt{\frac{1}{n}+1}$, so $f_{1}(z)=\sqrt{z+1}$ and $f_{2}(z)=-\sqrt{z+1}$. Now I have something to do with the identity theorem and accumulation points, but I'm not sure how this works... If someone could help me with this problem, that would be appreciated!
Note that $\frac n{n+1}=\frac1{1+1/n}$. There is an analytic function $f\colon\mathbb{D}\longrightarrow\mathbb C$ satisfying your condition if and only if$$(\forall z\in\mathbb{D}):f^2(z)=\frac1{1+z},\tag1$$by the identity theorem. Consider $\log\colon\{z\in\mathbb{C}\,|\,|z-1|<1\}\longrightarrow\mathbb C$ defined by$$\log z=z-1-\frac{(z-1)^2}2+\frac{(z-1)^3}3+\cdots$$The you can take $f(z)=\pm\exp\left(\frac12\log(z+1)\right)$ and these are the only solutuions to your problem.
But if you replace $\mathbb D$ with $\mathbb C$, then there is no solution, because it follows from $(1)$ that $\lim_{z\to-1}f(z)$ does not exist.