I have the following problem:
Let $D \subset \mathbb{C}$ denote the unit disk and let $S$ denote the strip $\{z \in \mathbb{C} : |\operatorname{Im}z| < \pi/2\}$.
Find a conformal map $f$ from $D$ onto $S$ satisfying $f(0) = 0$ and $f'(0) = 2$.
Here's the process I've taken so far.
First, I know that we have a biholomorphism from the right half plane to the unit disk given by $\frac{z-1}{z+1}$. Taking the inverse of this, we get that $\frac{-1-z}{z-1}$ is a biholomorphism from the unit disk to the right half plane.
Now, we have a biholomorphism from the right half plane to the strip $\{z \in \mathbb{C} \mid |\operatorname{Im}z| < 1 \}$ given by $e^{-z}$.
Finally, a biholomorphism from $\{z \in \mathbb{C} \mid |\operatorname{Im}z| < 1 \}$ to $\{z \in \mathbb{C} \mid |\operatorname{Im}z| < \pi/2 \}$ is simply $(\pi/2)z$.
Composing these gives us a biholomorphism from the unit disk to $S$ given by $(\pi/2)e^{\frac{z+1}{z-1}}$.
Now, however, when I plug in 0 to this, I get $(\pi/2)e^{-1} \neq 0$. So, my logic is the following, if we simply subtract $(\pi/2)e^{-1}$, this is a biholomorphism on $S$ since it doesn't change the imaginary part, and subtracting a constant is clearly holomorphic and bijective.
So then we have $f(z) = (\pi/2)e^{\frac{z+1}{z-1}} - (\pi/2)e^{-1}$ is a biholomorphism from $D$ to $S$ that satisfies $f(0) = 0$. Now, I just want to get that $f'(0) = 2$.
When I differentiate, I get $f'(z) = (\pi/2)e^{\frac{z+1}{z-1}} (\frac{(z-1)-(z+1)}{(z-1)^2}) = - \pi e^{\frac{z+1}{z-1}}$, so $f'(0) = -\pi e^{-1}$. I'm not quite sure where to go from here. Part of the issue in my thinking (if anyone could clarify this) is the following. If I multiply this by $-1$, then wouldn't I have a biholomorphism from $D$ to $S$ such that $f(0) = 0$ and $f'(0) = \pi e^{-1} > 0$? I thought that the Riemann Mapping Theorem implied that if we had a biholomorphism that satisfied these properties ($f(0) = 0$ and $f'(0) > 0$), then it was unique. If this is true, that would mean that there is no way to get a biholomorphism from $D$ to $S$ such that $f'(0) = 2$ since it must be equal to $\pi e^{-1}$.
Any insights on this would be appreciated!
The condition $|f'(0)|=2$ is superfluous: once you've found such a map, the value of $|f'(0)|$ is determined and can't be changed anyway. It will work out correctly in this case, though. reuns is correct that you don't really have the behaviour of $e^{-z}$ correct, it doesn't take a half-plane conformally to a strip but rather the reverse, so it might be easier to try to do the reverse map (the strip to the disk) and then invert it.