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Complex Function

Conformal Mapping

Intuitively, a conformal mapping is a "mapping that preserves angles between curves". In other words. when defining curves, the angles between curves must also be defined.

Paths

By definition. A path in the complex plane from a point A to a point B is a continuous function 𝛾:[𝑎,𝑏]→ℂ such that 𝛾(𝑎)=𝐴 and 𝛾(𝑏)=𝐵.

Examples

• Example 1

  𝛾(𝑡)=(2+𝑖)+ℯ𝑖𝑡, 0≤𝑡≤𝜋  =(2+cos𝑡) 𝑢(𝑥,𝑦) +𝑖(1+sin𝑡) 𝑦(𝑡)

• Example 2

  𝛾(𝑡)=(2+𝑖)+𝑡(-3-5𝑖), 0≤𝑡≤1  =(2-3𝑡)+𝑖(1−5𝑡)

• Example 3

  𝛾(𝑡)=𝑡ℯ𝑖𝑡, 0≤𝑡≤3𝜋  =(𝑡cos𝑡)+𝑖(𝑡sin𝑡)

• Example 4

  𝛾(𝑡)={𝑡(1+𝑖), 0≤𝑡≤1 𝑡+𝑖, 1≤𝑡≤2 2+𝑖(3-𝑡), 2≤𝑡≤3

Curves

By definition. A path 𝛾:[𝑎,𝑏]→ℂ is smooth if the functions 𝑥(𝑡) and 𝑦(𝑡) in the representation 𝛾(𝑡)=𝑥(𝑡)+𝑖𝑦(𝑡) are smooth, that is, have as many derivatives as desired.

In the above examples, (1), (2), and (3) are smooth, whereas (4) is piecewise smoth, i.e. put together ("concateated") from finitely many smooth paths.

The term curve is typically used for a smooth or piecewise smooth path.

If 𝛾=𝑥+𝑖𝑦:[𝑎,𝑏]→ℂ is a smooth curve and 𝑡₀∈(𝑎,𝑏), then

𝛾′(𝑡0)=𝑥′(𝑡0)+𝑖𝑦′(𝑡0)

is a tangent vector to 𝛾 at 𝑧₀=𝛾(𝑡₀)

The Angle between Curves

By definition. Let 𝛾₁ and 𝛾₂ be two smooth curves, intersecting at a point 𝑧₀. The angle between the two curves at 𝑧₀ is defined as the angle between the two tangent vectors at 𝑧₀.

Example

Let 𝛾₁:[0,𝜋]→ℂ, 𝛾₁(𝑡)=ℯ^𝑖𝑡 and

𝛾₂:[𝜋2,3𝜋2)→ℂ, 𝛾₂(𝑡)=2+𝑖+ℯ^𝑖𝑡.

Then 𝛾₁(𝜋2)=𝛾₂(𝜋)=𝑖. Furthermore,

𝛾′1(𝑡)=𝑖ℯ𝑖𝑡, 𝛾′1(𝜋2)=𝑖ℯ𝑖𝜋2=𝑖2=−1 𝛾′2(𝑡)=2𝑖ℯ𝑖𝑡, 𝛾′2(𝜋)=2𝑖ℯ𝑖𝜋=2𝑖(−1)=−2𝑖

The angle between these curves at 𝑖 is thus 𝜋2.

Conformality

By definition. A function is conformal if it preserves angles between curves. More precisely, a smooth complex-valued function 𝑔 is conformal at 𝑧₀ if whenever 𝛾₁ and 𝛾₂ are two curves that intersect at 𝑧₀ with non-zero tangents, the 𝑔∘𝛾₁ and 𝑔∘𝛾₂ have non-zero tangents at 𝑔(𝑧₀) that intersect at the same angle.

A conformal mapping of a domain 𝐷 onto 𝑉 is a continuously differentiable mapping that is conformal at each point in 𝐷 and maps 𝐷 one-to-one onto 𝑉.

Analytic Functions

By theorem. If 𝑓:𝑈→ℂ is analytic and if 𝑧₀∈𝑈 such that 𝑓′(𝑧₀)≠0, then 𝑓 is conformal at 𝑧₀.

Reason: If 𝛾:[𝑎,𝑏]→𝑈 is a curve in 𝑈 with 𝛾(𝑡₀)=𝑧₀ for some 𝑡₀∈(𝑎,𝑏), then

(𝑓∘𝛾)′(𝑡0)=𝑓′(𝛾(𝑡0))⋅𝛾′(𝑡0)=𝑓′(𝑧0) ∈ℂ\{0} ⋅𝛾′(𝑡0)

Thus (𝑓∘𝛾)′(𝑡₀) is obtained from 𝛾′(𝑡₀) via multiplication by 𝑓′(𝑧₀) (=rotation & stretching).

If 𝛾₁, 𝛾₂ are two curves in 𝑈 through 𝑧₀ with tangent vectors 𝛾′₁(𝑡₁), 𝛾′₂(𝑡₂), then (𝑓∘𝛾₁)′(𝑡₁) and (𝑓∘𝛾₂)′(𝑡₂) are both obtained from 𝛾′₁(𝑡₁), 𝛾′₂(𝑡₂), respectively, via multiplication by 𝑓′(𝑧₀). The angle between them is thus preserved.

Example

• 𝑓(𝑧)=𝑧² maps 𝑈={𝑧∈ℂ|Re𝑧>0} conformally ont ℂ\(−∞,0].
• 𝑓(𝑧)=ℯ^𝑧 is conformal at each point in ℂ (𝑓 is analytic in ℂ and 𝑓′(𝑧)≠0 in ℂ. Since 𝑓 is not one-to-one in ℂ, it is not a conformal mapping from ℂ onto ℂ\{0}.

  However, if you choose 𝐷={𝑧|0<Im𝑧<2𝜋}, then 𝑓 maps 𝐷 conformally onto 𝑓(𝐷)=ℂ\[0,∞).

• 𝑓(𝑧)=𝑧 is one-to-one and onto from ℂ to ℂ, however, angles between curves are reversed in orientation. 𝑓 is thus not conformal anywhere.