I am trying to find a conformal map $f(z)$ from the region $A=\{z:|z|<1, Im(z)>\frac{1}{2}\}$ to the unit disc $\mathbb{D}$. Namely, the upper $\frac{1}{4}$ disc to the unit disc.

I am really stuck in this question.

I firstly tried some conformal translation map, so that I could kind of "extended" the area, but I failed.

Any hints or explanations are really appreciated!!!


Note that $$ A=\left\{z\in\mathbb{C}:\Im\left(z\right)>\frac{1}{2}\right\}\cap\mathbb{D}, $$ i.e., $A$ is the intersection of two disks (in complex analysis, a half-plane is also regarded as a disk, with its radius being infinity). Further, the intersection of the two circles is $$ \left\{\frac{\sqrt{3}}{2}+\frac{i}{2},-\frac{\sqrt{3}}{2}+\frac{i}{2}\right\}=\left\{z\in\mathbb{C}:\Im\left(z\right)=\frac{1}{2}\right\}\cap\partial\mathbb{D}. $$

To figure out a conformal mapping $f$ from $A$, or similar disk-disk intersections, to $\mathbb{D}$, a systematical way follows these three steps.

  • Step 1: Maps $A$ to a wedge-like domain $B=\left\{z\in\mathbb{C}:\arg\left(z\right)\in\left(0,\alpha\right)\right\}$ for some $\alpha\in\left(0,\pi\right)$ using a Mobius transform.
  • Step 2: Maps $B$ to the upper half-plane $\mathbb{H}$ using a power function.
  • Step 3: Maps $\mathbb{H}$ to $\mathbb{D}$ using another Mobius transform.

Let us figure out these three steps.

The first step is crucial; we need to realize why a Mobius transform suffices to map $A$ to some wedge-like domain. Well, on the one hand, a Mobius transform maps (part of) circles to (part of) circles (again, a straight line in complex analysis is also regarded as a circle, with its radius being infinity). On the other hand, a Mobius transform, as a conformal mapping, maps the boundary of a domain to the boundary of its image. With this understanding, if we choose to map $\left(-\sqrt{3}+i\right)/2$ to $0$, and $\left(\sqrt{3}+i\right)/2$ to $\infty$, both of the boundaries must become straight lines (because the two boundaries, as part of two circles, intersect at $\left(\pm\sqrt{3}+i\right)/2$; after transformation, the image of these two boundaries must also be part of two circles, intersecting at $0$ and $\infty$; only two straight lines could make it), making the image of the domain wedge-like.

Therefore, consider ($C\in\mathbb{C}$ is a constant) $$ f_1(z)=C\frac{2z-\left(-\sqrt{3}+i\right)}{2z-\left(\sqrt{3}+i\right)}, $$ which maps $\left(-\sqrt{3}+i\right)/2$ to $0$, and $\left(\sqrt{3}+i\right)/2$ to $\infty$. In addition, we hope to make one side of the wedge coincide with the positive real axis, i.e., $$ f_1\Bigl(x+\frac{1}{2}i\Bigr)\in\mathbb{R}^+ $$ for all $x\in\left(-\sqrt{3}/2,\sqrt{3}/2\right)$. This requires $C\in\mathbb{R}^-$. Thus without loss of generality, take $C=-1$ and we have $$ f_1(z)=-\frac{2z-\left(-\sqrt{3}+i\right)}{2z-\left(\sqrt{3}+i\right)}. $$ Further, since the angle formed between $\left\{z\in\mathbb{C}:\Im\left(z\right)=1/2\right\}$ and $\partial\mathbb{D}$ is $\pi/3$ (and a conformal mapping always preserves the angle in its holomorphic domain), we have $$ f_1:A\to B=\left\{z\in\mathbb{C}:\arg\left(z\right)\in\left(0,\frac{\pi}{3}\right)\right\}. $$

The second step is rather simple, $$ f_2(z)=z^3 $$ suffices to map $B$ to $\mathbb{H}$.

The last step is also standard, $$ f_3(z)=\frac{z-i}{z+i} $$ suffices to map $\mathbb{H}$ to $\mathbb{D}$.

To sum up, $$ f=f_3\circ f_2\circ f_1:A\to\mathbb{D} $$ would be a candidate for the expected conformal mapping.

  • $\begingroup$ WOW!!! Thank you so much $\endgroup$ – JacobsonRadical Apr 29 '18 at 4:41
  • $\begingroup$ @JacobsonRadical: My pleasure! You may also explore to map $B(p_1,r_1)\cap B(p_2,r_2)$ to $\mathbb{D}$, where $B(p,r)$ denotes the disk centered at $p\in\mathbb{C}$ with radius $r\in\left(0,\infty\right]$, as long as $B(p_1,r_1)\cap B(p_2,r_2)\ne\emptyset$. $\endgroup$ – hypernova Apr 29 '18 at 4:48

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.