# Find a conformal map from the upper $\frac{1}{4}$ disc to the whole unit disc

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.

• WOW!!! Thank you so much – JacobsonRadical Apr 29 '18 at 4:41
• @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 $B(p_1,r_1)\cap B(p_2,r_2)\ne\emptyset$. – hypernova Apr 29 '18 at 4:48