Is there a solution of this complex analysis problem? The problem is to find a linear fractional transformation $w(z)$ that transforms the circle $|z|<2$ into the semiplane ${\rm Re}(w)>0$ and satisfies $w(0)=1$ and $w'(0)= \pi/2$.
For those who do not know, a linear fractional transformation is a transformation of the form
$$
w(z) = \frac{a z + b}{c z + d},
$$
where $a$, $b$, $c$, and $d$ are complex numbers.
A friend of mine was given this problem as a part of an assignment, and her calculations indicate that there is no transformation satisfying the problem's requirements. Unable to find a logical or arithmetic mistake in her calculations and highly doubting that the problem has no solution, she asked me to post a question here about the problem as she isn't proficient in English herself.
Is there a solution of this problem? If so, what are the coefficients?
 A: It turned out that the lecturer had mistyped the problem. Instead of the condition $w'(0) = \pi/2$, he should have typed ${\rm arg} [w'(0)]=\pi/2$.
The problem as typed has no solution indeed, so my friend was right.
Unfortunately, students have to pay with their time and sleepless nights for such carelessness of some math teachers. That's something all math teachers have to always remember!

UPDATE: Here's a proof that the problem has no solution.
As stated in the post with the question, a linear fractional transformation is a transformation of the form
$$
w(z) = \frac{a z + b}{c z + d},
$$
where $a$, $b$, $c$, and $d$ are complex numbers.
Let's assume there's a linear fractional transformation that satisfies the problem's requirements. Our goal is to arrive at a contradiction.
The first observation is that the coefficient $d$ must be non-zero, as otherwise $w(0)$ would be $\infty$, while the problem states that $w(0)=1$.
Using the fact that $d$ is non-zero, we normalize all coefficients by dividing them by $d$. The transformation doesn't change and $d$ becomes unity:
$$
w(z) = \frac{a z + b}{c z + 1}.
$$
The conditions $w(0)=1$ and $w'(0)=\pi/2$ can now be written as follows:
$$
b = 1,
$$
$$
a - c = \pi/2.
$$
Let's now require that the circle $|z|=2$ become the straight line ${\rm Re}(w)=0$. Any point $z$ on the circle $|z|=2$ can be expressed as $z = 2\exp(i \phi)$, where $\phi$ is a real number, and any point $w$ on the line ${\rm Re}(w)=0$ can be expressed as $w = i \psi$, where $\psi$ is a real number. Substituting this into $w(z)$ and using the fact that $b=1$, we get
$$
\frac{2a\exp(i\phi)+1}{2c\exp(i\phi)+1} = i \psi.
$$
Let's now use the above equation to express $\exp(i\phi)$ in terms of $\psi$:
$$
\exp(i\phi) = \frac{1 - i\psi}{2ic\psi - 2a}.
$$
We will now use the fact that for any real $\psi$, the right-hand side of this equation must return a complex number whose absolute value is unity, for the absolute value of the left-hand side is always unity for any real $\phi$.
To use the above fact, we focus on two particular cases: (a) $\psi \to \infty$ and (b) $\psi = 0$. Considering case (a), we get
$|-1/2c|=1$, which means $|c|=1/2$. Considering case (b), we get $|-1/2a|=1$, which means $|a|=1/2$.
Let's now summarize what we know about $a$ and $c$:
(1) $|a|=1/2$,
(2) $|c|=1/2$,
(3) $a-c=\pi/2$.
Now it's easy to see that conditions (1)-(3) can't be satisfied simultaneously. Indeed, conditions (1) and (2) require that $a$ and $c$ be on the circle centered at the center of the complex plane and with a radius of $1/2$, while condition (3) requires that $c$ be to the left from $a$ at a distance of $\pi/2$ from it. No two points of the above circle are at such a distance from each other, since its diameter is 1, which is smaller than $\pi/2$.
Thus we have arrived at a contradiction, and hence there's no linear fractional transformation satisfying the problem's requirements.

UPDATE 2: If the condition $w'(0)=\pi/2$ is replaced by what the lecturer should have typed instead, namely ${\rm arg }[w'(0)]=\pi/2$, the transformation satisfying the requirements is:
$$
w(z)= -\frac{z-2i}{z+2i}.
$$
