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Complex Analysis

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Complex Function
 Conformal Mapping
 The Angle between Curves
  Analytic Functions


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.


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.


  • 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


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


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.


Let 𝛾1:[0,𝜋]→ℂ, 𝛾1(𝑡)=ℯ𝑖𝑡 and

𝛾2:[𝜋2,3𝜋2)→ℂ, 𝛾2(𝑡)=2+𝑖+ℯ𝑖𝑡.

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.


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 𝛾1(𝑡1), 𝛾2(𝑡2), then (𝑓∘𝛾1)′(𝑡1) and (𝑓∘𝛾2)′(𝑡2) are both obtained from 𝛾1(𝑡1), 𝛾2(𝑡2), respectively, via multiplication by 𝑓′(𝑧0). The angle between them is thus preserved.


  • 𝑓(𝑧)=𝑧2 maps 𝑈={𝑧∈ℂ|Re𝑧>0} conformally ont ℂ\(−∞,0].
  • 𝑓(𝑧)=ℯ𝑧 is conformal at each point in (𝑓 is analytic in and 𝑓′(𝑧)≠0 in . Since 𝑓 is not one-to-one in , it is not a conformal mapping from 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 to , however, angles between curves are reversed in orientation. 𝑓 is thus not conformal anywhere.


ID: 190400014 Last Updated: 2019/4/14 Revision:


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