# Complex Map From the Unit Disc to the Right Half Plane

So I have a map from the unit disc to the upper half plane:

$$f(z) = \frac{z+i}{iz+1}$$

My logic is to now rotate this map by multiplying by $$-i$$ to arrive at the right half plane - though this does not seem to be a correct answer - what is the mistake I am making here?

Thanks

• What do you get as the final result, and why do you think that it is not correct? Commented Jan 28, 2021 at 12:28
• I've mapped this using an online complex mapper and the second part of the question I'm attempting uses the function derived to produce the Koebe function by dilating by 1/2 - squaring then taking away a quarter - I can't seem to derive said Koebe function this way. Commented Jan 28, 2021 at 12:31
• And as a final result I get $\frac{-iz+1}{iz+1}$ Commented Jan 28, 2021 at 12:32
• Your mapping to the upper half plane looks already wrong. It should be something like $f(z) = i \frac{1+z}{1-z}$. Commented Jan 28, 2021 at 12:33
• Crikey! Thankyou for spotting this as it's in the lecture notes - I shall try your suggested function and let you know! Commented Jan 28, 2021 at 12:34

I find it simpler to remember what the conformal mappings from a half plane to a disk are. As an example, the right half plane is the locus of points which are closer to $$+1$$ than to $$-1$$, i.e. the points $$w$$ for which $$|w-1| < |w+1|$$. It follows that $$T(w) = \frac{w-1}{w+1}$$ is the Möbius transformation which maps the right half plane onto the unit disk. Now you can solve the equation $$z = \frac{w-1}{w+1}$$ for $$w$$ to get the inverse mapping $$S(z) = T^{-1}(z) = \frac{1+z}{1-z}$$ which maps the unit disk onto the right half plane.
This is related to your result via $$-i f(z) = -i \frac{z+i}{iz+1} = \frac{-iz+1}{iz+1} = S(-iz)$$ which is therefore correct as well: $$-if$$ is a rotation by $$-i$$ which maps the unit disk onto itself, followed by $$S$$ which maps the unit disk onto the right half plane.