Provide injection holomorphic such, that $ \varphi(U)=\{z \in \mathbb{C}: \operatorname{Re} z>0, \operatorname{Im} z>0\} $ 
Let $U=B(1+i, \sqrt{2}) \cap B(1-i, \sqrt{2}) .$ Please provide
   injection holomorphic function $\varphi: U \rightarrow \mathbb{C}$
   such, that $$ \varphi(U)=\{z \in \mathbb{C}: \operatorname{Re} z>0,
 \operatorname{Im} z>0\} $$ (we have infinitely many of this function
   compute only one).

I am a bit confusing. I tried compute the Möbius transformation.
$f(z)=\frac{a z+b}{c z+d}$
$\varphi(0) = 0 = =\frac{b}{d} = 0 \iff b = 0$
$\varphi(1) = \frac{a}{c+d} = \infty \iff c+d = 0 \iff c = -d$
But I have problem with the second points, because we have collinear points. 
It's good direction of solution? 
 A: The given region $U$ is bounded by two circular arcs which terminate at $0$ and $2$ in the complex plane. Suppose we choose a Mobius transformation $$f(z) = \frac{az+b}{cz+d}$$ sending $f(0) = 0$ and $f(2) = \infty$. This implies $b = 0$ and $2c+d = 0$, so all such Mobius transformations are of the form $$f(z) = \frac{\lambda z}{z-2}$$ for some $\lambda \in \mathbb{C}$. Finally, choose $\lambda$ so that $f(1)=-\lambda$ is in the first quadrant, with $\text{Re}(z) > 0$ and $\text{Im}(z) > 0$, e.g. by taking $\lambda = -(1+i)$. 
Note that $U$ can be exhausted by circular arcs joining $0$ and $2$ (corresponding to circles through $0$ with center on the line $\text{Re}(z) = 1$). Our map $f$ sends each of these circles to a ray emanating from $0$. For example, it sends the "circular arc" $\text{Im}(z) = 0$ to the ray $\text{Re}(z) = \text{Im}(z)$ (i.e., the part of this line in the first quadrant).
Where does $f$ send $$\partial U = \{\text{Arc containing $1+i$}\} \cup \{\text{Arc containing $1-i$}\}?$$ We can calculate $f(1+i) = -1+i$ and $f(1-i) = 1-i$ to see that $f$ sends $\partial U$ to the line $\text{Re}(z) = -\text{Im}(z)$, sending the top arc to the ray in the second quadrant and the bottom arc to the ray in the fourth quadrant. We conclude that $f$ sends $U$ biholomorphically to a half-plane. 
We can now post-compose with a rotation to make this image the upper half-plane, and then a (suitable branch of the) square root map to make it the first quadrant, as desired. More precisely, let $g(z) = e^{2\pi i/8} z$ and $h(z) = \sqrt{z}$, taking the standard branch with cut along the positive real axis. Then, $h \circ g \circ f$ is a biholomorphism from $U$ to $\{z\in \mathbb{C} \, : \, \text{Re}(z) > 0 \text{ and } \text{Im}(z) > 0\}$.
