Does there exist a linear fractional transformation $w$ such that maps the region $\{z;\Re z>0\}$ onto the region $\{w;\Im w>0\}$ in such a way that $w(1)=i$ and $\arg w'(1)=\frac{\pi}3$?
I think the general form of this map is $w=az$ as we should have $w(0)=0$ and $w(\infty)=\infty$. So it follows from $w(1)=i$ that $w=iz$, but clearly $\arg w'(1)\neq\frac{\pi}3$? So probably my textbook is wrong!
The general form of a linear fractional transformation mapping the real axis to itself is $(a z + b)/(c z + d)$ with $a, b, c, d$ real. The composition with rotation by $\pi/2$ gives $(i a z + b)/(i c z + d)$ for a transformation that takes the imaginary axis to the real axis. Using the condition $w(1) = i$ and assuming $a \neq 0$, we can simplify the general form to $(i z - a)/(i a z + 1), \;a \in \mathbb R$.
For
amusementeducational purposes, let's find $a$ geometrically. $w(-1/(i a)) = \infty$, therefore $w(1/(i a)) = x_c = (1/a - a)/2$ is the center of the circle to which $w$ maps the real line. Then $\pi/3$ is the angle between the real line and the tangent to the circle at $i$, that is, $x_c = \tan(\pi/3)$. The images of $-1, 0, 1$ are $-i, -a, i$, which have to be traversed in the clockwise order. This gives $a > 0$, leaving one of the two roots.