Find the map from $\{ z: - \pi/2 < Im(z)<\pi/2\}$ to the vertical strip $\{ z: 0 < Re(z)<\log 2\}$.

Question

Find the map from $\{ z: - \pi/2 < Im(z)<\pi/2\}$ to the vertical strip $\{ z: 0 < Re(z)<\log 2\}$.

Using the map $f(z)=i (2/\pi)(\log 2) z$

we get the image of $f$ as $\{ z: \log 1/2 < Re(z)<\log 2\}$. But not the required one. How can I get that? Thanks

Answer 1

You are on the right track. I believe you start with the mapping $ z \rightarrow iz$, which maps the horizontal strip to the corresponding vertical strip.

Then use $ \zeta \rightarrow \zeta+\pi/2$, which moves the vertical strip in the positive right direction.

Finally use $\omega \rightarrow \frac{\omega log(2)}{\pi}$.

Now compose all of these together.

Answer 2

1. Shift the horizontal strip upwards by $\pi/2$ units ($+i\pi/2$), so that the strip lies on the real axis.
2. Rotate the strip by $\pi/2$ clockwise ($\cdot (-i)$)
3. Adjust the width of the strip by multiplying by a factor of $\log 2/\pi$.

This gives $f(z) = \dfrac{\log2}{\pi} \, (-i) \left(z + i \, \dfrac\pi2\right)$.

Answer 3

$f(z)=\frac{az+b}{cz+d}$.

Let's put $f(-\frac{\pi}2i)=0, f(\frac{\pi}2i)=\log2$ and $f(0)=\frac{\log2}2$.

So, $a=-\frac2{\pi}ib$. And $\frac{b+b}{\frac{\pi}2ic+d}=\log2 $. Finally, $b=\frac{\log2}2d\implies c=0$.

Putting this together, $f(z)=\frac{-\frac2{\pi}biz+b}{\frac2{\log2}b}$, or $\boxed{f(z)=-\frac{\log2}{\pi}(iz-\frac{\pi}2)}$.