Must a holomorphic function from $D(0,1)$ to $D(0,1)$ have a fixed point? Must every holomorphic function $f:D(0,1)\longrightarrow D(0,1)$ have a fixed point? 
I know that any holomorphic function with two fixed points is the identity: $f=Id$, but I can't find out an holomorphic function without a fixed point. 
Appreciate any suggestion.
 A: It depends on the precise definition of $D(0,1)$. If this is open disk then the answer is negative. Indeed, open disk $D$ is conformal to the upper half plane $U$, say, via the map $g: D\to U$. In the latter take the map $f(z)=z+1$ which has no fixed points in the half-plane. Thus, the holomorphic self-map $F=g^{-1}\circ f\circ g $ of $D$ has no fixed points in $D$ either. 
If by a disk you mean a closed disk then every continuous self-map of the closed disk (of any dimension) has a fixed point. This is Brouwer's fixed point theorem. 
A: If your domain $\mathbb{D}$ is the unit disk then your map $f$ is a conformal self-map. If it's not the identity map then it either has two fixed points on the boundary (counting multiplicity) or one fixed point inside the disk. This follows from the fact that You can write the equation for conformal self-map and make it equal to its fixed point. Then you will find a quadratic equation whose product of roots has modulus $1$.
A: Need not be. 
Counterexamples: $f(z)=(z+1)/2$
A: The answer is no. We know that the unit disk $\mathbb{D}$ is conformally equivalent to the upper half-plane $\mathbb{H}$. Consider any conformal map sending $\mathbb{D} \rightarrow \mathbb{H}$. Then consider the map $z \rightarrow z+1$ which certainly does not have any fixed points. Then take the inverse of the first map to get from $\mathbb{H}$ back to $\mathbb{D}$. This map does not have any fixed points.
As an example: $\displaystyle f(z) = i\frac{1+z}{1-z}$ sends $\mathbb{D}$ to $\mathbb{H}$. Then $g(z) = z+1$ is conformal and has no fixed points. Then the inverse of the above map, $\displaystyle f^{-1}(z) = \frac{z-i}{z+i}$ maps $\mathbb{H}$ back to $\mathbb{D}$. Composing these gives: $\displaystyle \frac{1-(1-2i)z}{1+2i-z}$ is a conformal automorphism of the disk with no fixed points.
