which mobius transformation maps the unit disc onto te right half plane?

Question

Which of the following mobius transformation maps the unit disc onto the right half plane.

a) $f(z)=\frac{z-i}{z+i}$

b)$f(z)=\frac{z-1}{z+1}$

c) $f(z)=\frac{1+z}{1-z}$

d)$f(z)=i\Big(\frac{1+z}{1-z}\Big)$

What is the bestway to deal with such questions in less time. I tried putting values for $z$ as 1 or i, but it did not work.

And one more thing, If I see this functions are not even defined for certain points on the unit circle. For example, option a , is not defined for $z=i$, option b not defined for $z=1$ and c and d are not defined for $z=1$. So how can this be even mapping from unit disk, as it is not taking some points on the unit disk.

Thanks in advance!

Answer 1

In terms of the cases that you mention for $z=1$ and $z=i$, usually when people talk about the unit disc, they are referring to the open disc with radius 1, which does not contain $z=1$ or $z=i$. In this case, a good test case is $z=0$, which is in fact contained in the unit disc. For a), $f(0) = -1$, so a) cannot be correct. For b), $f(0) = -1$, so b) cannot be correct either. For c), $f(0) = 1$, so c) could be correct. For d), $f(0) = i$; so d) cannot be correct. Therefore, c) must be correct.

The way that I usually think about problems involving conformal mapping to/from the unit disc is to express the other region in some way which makes it easy to design a Möbius map from that region to the unit disc. In the case of the right half-plane, we have that if $z$ is in the right-half plane, then $|z+1| > |z-1|$, so then we have that $\left|\frac{z-1}{z+1}\right| < 1$, so $\frac{z-1}{z+1}$ is in the unit disc, so $g(z) = \frac{z-1}{z+1}$ is a Möbius map from the right-half plane to the unit disc. Then the inverse of $g$ is a Möbius map from the unit disc to the right-half plane. You can check that the inverse of $g$ is the map from part c). Hope this helps!

Answer 2

Remember, a Mőbius transformation is actually a (holomorphic) bijection on the Riemann sphere, the one-point compactification of the complex plane. So it maps to and from the "point at infinity" as well. On it all the lines are "generalized circles". The straight lines in the complex plane are the ones that go through the point at infinity.

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The Cayley transformation maps the unit disk onto the upper half plane. It's given by $$-i\dfrac {z+1}{z-1}.$$

You can see this because $1,i,-1$ (which determine the unit circle) map to $ \infty,-1,0$ (which determine the $x$-axis) respectively.

The origin can be used as a test point, and goes to $i.$

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Now all we need to do is rotate by $-\pi/2.$ That's multiply by $e^{\frac {-\pi i}2}=-i,$ to get $$\dfrac {1+z}{1-z}.$$