1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$
  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 $f(z) = \dfrac{\log2}{\pi} \, (-i) \left(z + i \, \dfrac\pi2\right)$.

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.