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

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

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.

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

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

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 $$\boxed{f(z)=-\frac{\log2}{\pi}(iz-\frac{\pi}2)}$$.