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Riemann Mapping Theorem

Conformal Mappings

Properties

• The conformal mappings from ℂ to ℂ are of the form 𝑧⟼𝑎𝑧+𝑏𝑐𝑧+𝑑.
• The conformal mappings from ℂ to ℂ are of the form 𝑧⟼𝑎𝑧+𝑏.

More is true

• There is no conformal mapping 𝑓:ℂ→𝐷, where 𝐷⊂ℂ, 𝐷≠ℂ.
• There is no conformal mapping 𝑓:ℂ→𝐷, where 𝐷⊂ℂ.

And what conformal mappings are ther of the form 𝑓:𝔻→𝐷, where 𝔻=𝐵₁(0) is the unit disk and 𝐷⊂ℂ?

The Riemann Mapping Theorem

By theorem. If 𝐷 is a simply connected domain (= open, connected, no holes) in the complex plane, but not the entire complex plane, then there is a conformal map (= analytic, one-to-one, onto) of 𝐷 onto the open unit disk 𝔻.

That is "𝐷 is conformally equivalent to 𝔻"

The Riemann Map

Let 𝐷 be a simply connected domain. In order to find a unique conformal mapping 𝑓 from 𝐷 onto 𝔻, "3 real parmeeters" are needed to specify. For example, specify

• a point 𝑧₀∈𝐷 that is to be mapped to 0 under 𝑓 (=2 real parameters 𝑥₀, 𝑥₁)
• the argument of 𝑓′(𝑧₀) (=1 real parameter_, for example by requiring that 𝑓′(𝑧₀)>0.

The Upper Half Plane

Let 𝐷 be the upper half plane, i.e. 𝐷={𝑧:Imz>0}. then 𝐷 can be mapped to 𝔻 via (the restriction of) a Mobius transformation: Let 𝑓 be the Mobius transformation that maps 0, 1, ∞ to 1, 𝑖, −1.

Then the line through 0, 1, ∞ (the real axis) must be mapped to the circle through 1, 𝑖, −1 (the unit circle). Further, the domain to the left of the real axis (𝐷) is then mapped to the domain to the left of the unit circle (𝔻), oriented by the ordering of the given points.

Finding the Riemann Map

The restriction of the Mobius transformation 𝑓 to the upper half plane 𝐷 thus maps 𝐷 onto 𝔻. Finding a formula for 𝑓, such that 𝑓 maps 0, 1, ∞ to 1, 𝑖, −1.

• 𝑓 is of the form 𝑓(𝑧)=𝑎𝑧+𝑏𝑐𝑧+𝑑. Since 𝑓(∞)≠∞, therefore 𝑐≠0 and thus assume 𝑐=1.
• Thus 𝑓(𝑧)=𝑎𝑧+𝑏𝑧+𝑑. Since 𝑓(∞)=−1, therefore 𝑎=−1.
• Then 𝑓(𝑧)=−𝑧+𝑏𝑧+𝑑. Since 𝑓(0)=1, therefore 𝑏𝑑=1, thus 𝑏=𝑑.
• So 𝑓(𝑧)=−𝑧+𝑏𝑧+b. Since 𝑓(1)=𝑖, therefore −1+𝑏1+b=𝑖, thus 𝑏=𝑖.

Thus 𝑓(𝑧)=−𝑧+𝑖𝑧+𝑖 maps the upper half plane 𝐷 conformally onto the unit disk 𝔻

The first quadrant to the upper half of the unit disk

Let 𝑄 be the first quadrant, i.e. the domain in the complex plane, bounded by the positive real axis and the positive imaginary axis. Since the map 𝑓 maps 0 to 1, 𝑖 to 0, and ∞ to −1, it maps the line through 0, 𝑖, ∞ (i.e. the imaginary axis) to the line through 1, 0, −1 (i.e. the real axis).

Hence the restriction of 𝑓 to 𝑄 maps 𝑄 conformally onto the upper half of the unit disk, 𝔻⁺

The first quadrant to the upper half plane

The map 𝑔(𝑧)=𝑧² is injective and analytic in the first quadrant 𝑄

𝑔 maps 𝑄 conformally onto its image, namely the upper half plane 𝐷

The Riemann Map of the upper half of the unit disk

The previous three examples help to construct the Riemann map from 𝔻⁺ to 𝐷:

• 𝔻⁺𝑓⁻¹⟼𝑄𝑔(𝑧)=𝑧²⟼ 𝐷𝑓(𝑧)=−𝑧+𝑖𝑧+𝑖⟼𝔻
• 𝔻⁺ℎ=𝑓∘𝑔∘𝑓⁻¹⟼𝔻

Application

Many problems are easier to solve in the unit disk (or some other "nice" standard region) than in the region they are formulated in.

Solutions can be found in the standard region, then transported back to the original region via a Riemann map

Example: Fluid flow can be modeled nicely in the upper half plane.

To understand a similar fluid flow in another region, map this flow from the upper half plane to the desired region using the Riemann map.

Other examples: electrostatics, heat conduction, aerodynamics, etc.