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The Cauchy-Riemann equations
โ€ƒComplex Function
โ€ƒCauchy-Riemann equations

source/reference:
https://www.youtube.com/channel/UCaTLkDn9_1Wy5TRYfVULYUw/playlists

The Cauchy-Riemann equations

Complex Function

A complex function ๐‘“ can be written as ๐‘“(๐‘ง)=๐‘ข(๐‘ฅ,๐‘ฆ)+๐‘–๐‘ฃ(๐‘ฅ,๐‘ฆ) where ๐‘ง=๐‘ฅ+๐‘–๐‘ฆ and ๐‘ข, ๐‘ฃ are real-valued functions that depend on the two real variables ๐‘ฅ and ๐‘ฆ.

Example:

๐‘“(๐‘ง)=๐‘ง2=(๐‘ฅ+๐‘–๐‘ฆ)2=๐‘ฅ2-๐‘ฆ2 ๐‘ข(๐‘ฅ,๐‘ฆ) +๐‘–โ‹…2๐‘ฅ๐‘ฆ ๐‘ฃ(๐‘ฅ,๐‘ฆ)

Thus ๐‘ข(๐‘ฅ,๐‘ฆ)=๐‘ฅ2-๐‘ฆ2 and ๐‘ฃ(๐‘ฅ,๐‘ฆ)=2๐‘ฅ๐‘ฆ.

For example, if ๐‘ง=2+๐‘–, then clearly ๐‘“(๐‘ง)=(2+๐‘–)2=4+4๐‘–+๐‘–2=3+4๐‘–.

Alternatively, since ๐‘ง=๐‘ฅ+๐‘–๐‘ฆ, then ๐‘ฅ=2 and ๐‘ฆ=1, thus

๐‘ข(๐‘ฅ,๐‘ฆ)=๐‘ข(2,1)=22-12=3 and ๐‘ฃ(๐‘ฅ,๐‘ฆ)=๐‘ฃ(2,1)=2โ‹…2โ‹…1=4

thus

๐‘“(2+๐‘–)=๐‘ข(2,1)+๐‘–๐‘ฃ(2,1)=3+4๐‘–

Another example

๐‘“(๐‘ง)=๐‘ง2=๐‘ข(๐‘ฅ,๐‘ฆ)+๐‘–=2๐‘ฃ(๐‘ฅ,๐‘ฆ), ๐‘ข(๐‘ฅ,๐‘ฆ)=๐‘ฅ2-๐‘ฆ2, ๐‘ฃ(๐‘ฅ,๐‘ฆ)=2๐‘ฅ๐‘ฆ

Therefore

  • ๐‘“ is differentiable everywhere in โ„‚
  • ๐‘“โ€ฒ(๐‘ง)=2๐‘ง for all ๐‘งโˆˆโ„‚

For the function ๐‘ข(๐‘ฅ,๐‘ฆ)

  • If fix the variable ๐‘ฆ at a certain value, then ๐‘ข only depends on ๐‘ฅ. For example, if only consider ๐‘ฆ=3, then ๐‘ข(๐‘ฅ,๐‘ฆ)=๐‘ข(๐‘ฅ,3)=๐‘ฅ2-9
  • This function can now differentiate with respect to ๐‘ฅ according to the rules of calculus and find that the derivative is 2๐‘ฅ
  • That is โˆ‚๐‘ขโˆ‚๐‘ฅ(๐‘ฅ,3)=𝑢๐‘ฅ(๐‘ฅ,3)=2๐‘ฅ, and, more generally, for arbitrary (fixed) ๐‘ฆ, โˆ‚๐‘ขโˆ‚๐‘ฅ(๐‘ฅ,๐‘ฆ)=๐‘ข๐‘ฅ(๐‘ฅ,๐‘ฆ)=2๐‘ฅ
  • This is called the partial derivative of ๐‘ข with respect to ๐‘ฅ

Similarly, for the function ๐‘ฃ(๐‘ฅ,๐‘ฆ)

  • For example, ๐‘ฃ(๐‘ฅ,3)=2โ‹…๐‘ฅโ‹…3=6๐‘ฅ, and the derivative of this function with respect to ๐‘ฅ is 6.
  • Thus โˆ‚๐‘ฃโˆ‚๐‘ฅ(๐‘ฅ,3)=๐‘ฃ๐‘ฅ(๐‘ฅ,3)=6
  • More generally, for arbitrary (fixed) ๐‘ฆ, โˆ‚๐‘ฃโˆ‚๐‘ฅ(๐‘ฅ,๐‘ฆ)=๐‘ฃ๐‘ฅ(๐‘ฅ,๐‘ฆ)=2๐‘ฆ
  • This is called the partial derivative of ๐‘ฃ with respect to ๐‘ฅ.

Obviously, the same thing can be done by fixing ๐‘ฅ and differentiating with respect to ๐‘ฆ.

  • Example: Let ๐‘ฅ =2. Then ๐‘ข(๐‘ฅ,๐‘ฆ)=๐‘ข(2,๐‘ฆ)=4โˆ’๐‘ฆ2 and โˆ‚๐‘ขโˆ‚๐‘ฆ(๐‘ฅ,๐‘ฆ)=๐‘ข๐‘ฆ(2,๐‘ฆ)=-2๐‘ฆ
  • More generally, โˆ‚๐‘ขโˆ‚๐‘ฆ(๐‘ฅ,๐‘ฆ)=๐‘ข๐‘ฆ(๐‘ฅ,๐‘ฆ)=-2๐‘ฆ
  • This is called the partial derivative of ๐‘ข with respect to ๐‘ฆ.
  • Similarly, ๐‘ฃ(2,๐‘ฆ)=4๐‘ฆ, and โˆ‚๐‘ฃโˆ‚๐‘ฆ(2,๐‘ฆ)=๐‘ฃ๐‘ฆ(2,๐‘ฆ)=4. More generally, โˆ‚๐‘ฃโˆ‚๐‘ฆ(๐‘ฅ,๐‘ฆ)=๐‘ฃ๐‘ฆ(๐‘ฅ,๐‘ฆ)=2๐‘ฅ
  • This is called the partial derivative of ๐‘ฃ with respect to ๐‘ฆ.

And the result are

๐‘“(๐‘ง)=๐‘ง2=๐‘ข(๐‘ฅ,๐‘ฆ)+๐‘–=2๐‘ฃ(๐‘ฅ,๐‘ฆ), ๐‘ข(๐‘ฅ,๐‘ฆ)=๐‘ฅ2-๐‘ฆ2, ๐‘ฃ(๐‘ฅ,๐‘ฆ)=2๐‘ฅ๐‘ฆ

The derivatives:

๐‘“โ€ฒ(๐‘ง)=2๐‘ง, ๐‘ข๐‘ฅ(๐‘ฅ,๐‘ฆ)=2๐‘ฅ, ๐‘ข๐‘ฆ(๐‘ฅ,๐‘ฆ)=-2๐‘ฆ, ๐‘ฃ๐‘ฅ(๐‘ฅ,๐‘ฆ)=2๐‘ฆ, ๐‘ฃ๐‘ฆ(๐‘ฅ,๐‘ฆ)=2x

Notice:

  • ๐‘ข๐‘ฅ=๐‘ฃ๐‘ฆ
  • ๐‘ข๐‘ฆ=-๐‘ฃ๐‘ฅ
  • ๐‘“โ€ฒ=๐‘ข๐‘ฅ+๐‘–๐‘ฃ๐‘ฅ =๐‘“๐‘ฅ =-๐‘–(๐‘ข๐‘ฆ+๐‘–๐‘ฃ๐‘ฆ) =-๐‘–๐‘“๐‘ฆ

Another Example


๐‘“(๐‘ง)=2๐‘ง3-4๐‘ง+1, where ๐‘ง=๐‘ฅ+๐‘–๐‘ฆ  =2(๐‘ฅ+๐‘–๐‘ฆ)3-4(๐‘ฅ+๐‘–๐‘ฆ)+1  =2(๐‘ฅ3+3๐‘ฅ2๐‘–๐‘ฆ+3๐‘ฅ๐‘–2๐‘ฆ2+๐‘–3๐‘ฆ3)-4๐‘ฅ-4๐‘–๐‘ฆ+1  =(2๐‘ฅ3-6๐‘ฅ๐‘ฆ2-4๐‘ฅ+1) ๐‘ข(๐‘ฅ,๐‘ฆ) +๐‘–(6๐‘ฅ2๐‘ฆ-2๐‘ฆ3-4๐‘ฆ) ๐‘ฃ(๐‘ฅ,๐‘ฆ)

Then

๐‘ข๐‘ฅ(๐‘ฅ,๐‘ฆ)=6๐‘ฅ2-6๐‘ฆ2-4๐‘ฃ๐‘ฅ(๐‘ฅ,๐‘ฆ)=12๐‘ฅ๐‘ฆ ๐‘ข๐‘ฆ(๐‘ฅ,๐‘ฆ)=-12๐‘ฅ๐‘ฆ๐‘ฃ๐‘ฆ(๐‘ฅ,๐‘ฆ)=6๐‘ฅ2-6๐‘ฆ2-4

Thus, ๐‘ข๐‘ฅ=๐‘ฃ๐‘ฆ and ๐‘ข๐‘ฆ=-๐‘ฃ๐‘ฅ

The derivatives:

๐‘“โ€ฒ(๐‘ง)=6๐‘ง2-4, where ๐‘ง=๐‘ฅ+๐‘–๐‘ฆ =6(๐‘ฅ+๐‘–๐‘ฆ)2-4  =(6๐‘ฅ2-6๐‘ฆ2-4)+12๐‘–๐‘ฅ๐‘ฆ  =๐‘ข๐‘ฅ(๐‘ฅ,๐‘ฆ)+๐‘–๐‘ฃ๐‘ฅ(๐‘ฅ,๐‘ฆ)=-๐‘–(๐‘ข๐‘ฆ(๐‘ฅ,๐‘ฆ)+๐‘–๐‘ฃ๐‘ฆ(๐‘ฅ,๐‘ฆ))=๐‘“๐‘ฅ(๐‘ง)=-๐‘–๐‘“๐‘ฆ(๐‘ง)

Cauchy-Riemann equations

By Theorem. Suppose that ๐‘“(๐‘ง)=๐‘ข(๐‘ฅ,๐‘ฆ)+๐‘–๐‘ฃ(๐‘ฅ,๐‘ฆ) is differentiable at a point ๐‘ง0. Then the partial derivatives ๐‘ข๐‘ฅ, ๐‘ข๐‘ฆ, ๐‘ฃ๐‘ฅ, ๐‘ฃ๐‘ฆ exist at ๐‘ง0, and satisfy there:

๐‘ข๐‘ฅ=๐‘ฃ๐‘ฆ and ๐‘ข๐‘ฆ=-๐‘ฃ๐‘ฅ

These are called the Cauchy-Riemann Equations. Also,

๐‘“โ€ฒ(๐‘ง0)=๐‘ข๐‘ฅ(๐‘ฅ0,๐‘ฆ0)+๐‘–๐‘ฃ๐‘ฅ(๐‘ฅ0,๐‘ฆ0)=๐‘“๐‘ฅ(๐‘ง0)  =-๐‘–(๐‘ข๐‘ฆ(๐‘ฅ0,๐‘ฆ0)+๐‘–๐‘ฃ๐‘ฆ(๐‘ฅ0,๐‘ฆ0))=-๐‘–๐‘“๐‘ฆ(๐‘ง0)

Method of Proof

For the difference quotient,

๐‘“(๐‘ง0+โ„Ž)-๐‘“(๐‘ง0)โ„Ž

whose limit as โ„Žโ†’0 must exist if ๐‘“ is differentiable at ๐‘ง0

  • Let โ„Ž approach 0 along the real axis only first, and then along the imaginary axis only next. Both times the limit must exist and both limits must be the same.
  • Equating these two limits with each other and recognizing the partial derivatives in the expressions yields the Cauchy-Riemann equations.

Another example

Let ๐‘“(๐‘ง)=๐‘ง=๐‘ฅ-๐‘–๐‘ฆ, then ๐‘ข(๐‘ฅ,๐‘ฆ)=๐‘ฅ and ๐‘ฃ(๐‘ฅ,๐‘ฆ)=-๐‘ฆ, so

๐‘ข๐‘ฅ(๐‘ฅ,๐‘ฆ)=1๐‘ฃ๐‘ฅ(๐‘ฅ,๐‘ฆ)=0 ๐‘ข๐‘ฆ(๐‘ฅ,๐‘ฆ)=0๐‘ฃ๐‘ฆ(๐‘ฅ,๐‘ฆ)=-1

Since ๐‘ข๐‘ฅ(๐‘ฅ,๐‘ฆ)โ‰ ๐‘ฃ๐‘ฆ(๐‘ฅ,๐‘ฆ) (while ๐‘ข๐‘ฆ(๐‘ฅ,๐‘ฆ)=-๐‘ฃ๐‘ฅ(๐‘ฅ,๐‘ฆ) for all ๐‘ง, the function ๐‘“ is not differentiable anywhere.

Recall: If ๐‘“ is differentiable at ๐‘ง0 then the Cauchy-Riemann equations hold at ๐‘ง0. However, if ๐‘“ satisfies the Cauchy-Riemann equations at a point ๐‘ง0 then does this imply that ๐‘“ is differentiable at ๐‘ง0? And what is the sufficient conditions for differentiability.

By theorem. Let ๐‘“=๐‘ข+๐‘–๐‘ฃ be defined on a domain ๐ทโŠ‚โ„‚. Then ๐‘“ is analytic in >๐ท if and only if ๐‘ข(๐‘ฅ,๐‘ฆ) and ๐‘ฃ(๐‘ฅ,๐‘ฆ) have continuous first partial derivatives on >๐ท that satisfy the Cauchy-Riemann equations.

Example: ๐‘“(๐‘ง)=๐‘’๐‘ฅcos๐‘ฆ+๐‘–๐‘’๐‘ฅsin๐‘ฆ. The

๐‘ข๐‘ฅ(๐‘ฅ,๐‘ฆ)=๐‘’๐‘ฅcos๐‘ฆ๐‘ฃ๐‘ฅ(๐‘ฅ,๐‘ฆ)=๐‘’๐‘ฅsin๐‘ฆ ๐‘ข๐‘ฆ(๐‘ฅ,๐‘ฆ)=โˆ’๐‘’๐‘ฅsin๐‘ฆ๐‘ฃ๐‘ฆ(๐‘ฅ,๐‘ฆ)=๐‘’๐‘ฅcos๐‘ฆ

Thus the Cauchy-Riemann equations are satisfied, and in addition, the functions ๐‘ข๐‘ฅ, ๐‘ข๐‘ฆ, ๐‘ฃ๐‘ฅ, ๐‘ฃ๐‘ฆ are continuous in โ„‚. Therefore, the function ๐‘“ is analytic in โ„‚, thus entire.


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ID: 190300030 Last Updated: 3/30/2019 Revision: 0


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