Bijective holomorphic map A bijective holomorphic map from unit disk to itself will be rotation? That mean $f(z)=e^{i\alpha}z$? How do I approach to solve  this problem?In addition I want to know how one can remember the conformal maps  which sends unit disk to upper half plane or conversely,and all possible known place to other places like that?
 A: For $w\in\mathbb{D}$, consider the following function defined on $\mathbb{D}$:
$$B_w(z) = \frac{z-w}{1-\overline{w}z}.$$
This function is called a "Blaschke factor." It defines a bijective holomorphism from $\mathbb{D}$ to $\mathbb{D}$, but it is not a rotation if $w\neq 0$. This demonstrates there are many such mappings that are not rotations.
To see this, first note that that by an easy computation, $B_{w}$ is its own inverse, so it is biholomorphic. To see it maps $\mathbb{D}\rightarrow \mathbb{D}$, just take the modulus and do the standard manipulations, noting that $|z|<1$ and $|w|<1$. 
However, if you make the assumption that $f(0)=0$, your guess is correct. All biholomorphic maps of the unit disk that fix the origin are rotations. We use the Schwartz lemma to show this. If you are not familiar with this, leave a comment below and I will edit this answer to include a proof and discussion.
By the lemma, we have $|f'(0)|\le 1$. Because $f^{-1}$ is another origin preserving biholomorphism, we also have $|f^{-1}$'$(0)|\le 1$. Because we know by the differentiation rule for inverse functions that $f'(0)=\frac{1}{f^{-1}\text{'}(0)}$, we must have $|f'(0)|=|f^{-1}\text{'}(0)|=1$, and considering the equality case in the Schwartz lemma, we see $f$ must be a rotation.
We know that $f^{-1}$ is differentiable. Then the chain rule gives
$$f(f^{-1}(x))=x\rightarrow f'(f^{-1}(x))=\frac{1}{f^{-1}\text{'}(x)}$$
and noting here that $f^{-1}(0)=0$ gives the result.
To answer the second part of your question, I suggest you use Google and the literature available to you. The magic search terms are "automorphism group of the disk" and "automorphism group of the upper half plane." This will tell you about all possible biholomorphisms of each space. Now, because the half-plane is conformally equivalent to the unit disk, knowing the automorphism group of the unit disk is enough to know of all the the maps from the unit disk to the upper half-plane.
