
Complex AnalysisComplex NumberTopologyFunctionSequences and LimitsIteration of FunctionComplex Derivative
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โซโฌโญโฎโฏโฐโฑโฒโณ โฅโฎโฏโฐโฑ โ โฒ โณ โด โ โ สน สบ โต โถ โท
๏น ๏น ๏น ๏น ๏ธน ๏ธบ ๏ธป ๏ธผ ๏ธ ๏ธ ๏ธฟ ๏น ๏ธฝ ๏ธพ ๏น ๏น ๏ธท ๏ธธ โ โ โด โต โ โ โ โก
โโโโโคโฆโฅโงโโโโโโโฒโผโโถโบโปโฒโณ โผโฝโพโฟโโโโโโ
โโ โโโโโโโโโโโโโโโณโฅขโฅฃโฅคโฅฅโฅฆโฅงโฅจโฅฉโฅชโฅซโฅฌโฅญโฅฎโฅฏ
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ContentThe Cauchy-Riemann equations
source/reference: The Cauchy-Riemann equationsComplex FunctionA complex function ๐ can be written as ๐(๐ง)=๐ข(๐ฅ,๐ฆ)+๐๐ฃ(๐ฅ,๐ฆ) where ๐ง=๐ฅ+๐๐ฆ and ๐ข, ๐ฃ are real-valued functions that depend on the two real variables ๐ฅ and ๐ฆ. Example: ๐(๐ง)=๐ง2=(๐ฅ+๐๐ฆ)2=
Thus ๐ข(๐ฅ,๐ฆ)=๐ฅ2-๐ฆ2 and ๐ฃ(๐ฅ,๐ฆ)=2๐ฅ๐ฆ. For example, if ๐ง=2+๐, then clearly ๐(๐ง)=(2+๐)2=4+4๐+๐2=3+4๐. Alternatively, since ๐ง=๐ฅ+๐๐ฆ, then ๐ฅ=2 and ๐ฆ=1, thus ๐ข(๐ฅ,๐ฆ)=๐ข(2,1)=22-12=3 and ๐ฃ(๐ฅ,๐ฆ)=๐ฃ(2,1)=2โ
2โ
1=4
thus ๐(2+๐)=๐ข(2,1)+๐๐ฃ(2,1)=3+4๐
Another example ๐(๐ง)=๐ง2=๐ข(๐ฅ,๐ฆ)+๐=2๐ฃ(๐ฅ,๐ฆ), ๐ข(๐ฅ,๐ฆ)=๐ฅ2-๐ฆ2, ๐ฃ(๐ฅ,๐ฆ)=2๐ฅ๐ฆ
Therefore
For the function ๐ข(๐ฅ,๐ฆ)
Similarly, for the function ๐ฃ(๐ฅ,๐ฆ)
Obviously, the same thing can be done by fixing ๐ฅ and differentiating with respect to ๐ฆ.
And the result are ๐(๐ง)=๐ง2=๐ข(๐ฅ,๐ฆ)+๐=2๐ฃ(๐ฅ,๐ฆ), ๐ข(๐ฅ,๐ฆ)=๐ฅ2-๐ฆ2, ๐ฃ(๐ฅ,๐ฆ)=2๐ฅ๐ฆ
The derivatives: ๐โฒ(๐ง)=2๐ง, ๐ข๐ฅ(๐ฅ,๐ฆ)=2๐ฅ, ๐ข๐ฆ(๐ฅ,๐ฆ)=-2๐ฆ, ๐ฃ๐ฅ(๐ฅ,๐ฆ)=2๐ฆ, ๐ฃ๐ฆ(๐ฅ,๐ฆ)=2x
Notice:
Another Example ๐(๐ง)=2๐ง3-4๐ง+1, where ๐ง=๐ฅ+๐๐ฆ
=2(๐ฅ+๐๐ฆ)3-4(๐ฅ+๐๐ฆ)+1
=2(๐ฅ3+3๐ฅ2๐๐ฆ+3๐ฅ๐2๐ฆ2+๐3๐ฆ3)-4๐ฅ-4๐๐ฆ+1
=
Then ๐ข๐ฅ(๐ฅ,๐ฆ)=6๐ฅ2-6๐ฆ2-4๐ฃ๐ฅ(๐ฅ,๐ฆ)=12๐ฅ๐ฆ
๐ข๐ฆ(๐ฅ,๐ฆ)=-12๐ฅ๐ฆ๐ฃ๐ฆ(๐ฅ,๐ฆ)=6๐ฅ2-6๐ฆ2-4
Thus, ๐ข๐ฅ=๐ฃ๐ฆ and ๐ข๐ฆ=-๐ฃ๐ฅ The derivatives: ๐โฒ(๐ง)=6๐ง2-4, where ๐ง=๐ฅ+๐๐ฆ =6(๐ฅ+๐๐ฆ)2-4
=(6๐ฅ2-6๐ฆ2-4)+12๐๐ฅ๐ฆ
=๐ข๐ฅ(๐ฅ,๐ฆ)+๐๐ฃ๐ฅ(๐ฅ,๐ฆ)=-๐(๐ข๐ฆ(๐ฅ,๐ฆ)+๐๐ฃ๐ฆ(๐ฅ,๐ฆ))=๐๐ฅ(๐ง)=-๐๐๐ฆ(๐ง)
Cauchy-Riemann equationsBy Theorem. Suppose that ๐(๐ง)=๐ข(๐ฅ,๐ฆ)+๐๐ฃ(๐ฅ,๐ฆ) is differentiable at a point ๐ง0. Then the partial derivatives ๐ข๐ฅ, ๐ข๐ฆ, ๐ฃ๐ฅ, ๐ฃ๐ฆ exist at ๐ง0, and satisfy there: ๐ข๐ฅ=๐ฃ๐ฆ and ๐ข๐ฆ=-๐ฃ๐ฅ
These are called the Cauchy-Riemann Equations. Also, ๐โฒ(๐ง0)=๐ข๐ฅ(๐ฅ0,๐ฆ0)+๐๐ฃ๐ฅ(๐ฅ0,๐ฆ0)=๐๐ฅ(๐ง0)
=-๐(๐ข๐ฆ(๐ฅ0,๐ฆ0)+๐๐ฃ๐ฆ(๐ฅ0,๐ฆ0))=-๐๐๐ฆ(๐ง0)
Method of Proof For the difference quotient,
whose limit as โโ0 must exist if ๐ is differentiable at ๐ง0
Another example Let ๐(๐ง)=๐ง=๐ฅ-๐๐ฆ, then ๐ข(๐ฅ,๐ฆ)=๐ฅ and ๐ฃ(๐ฅ,๐ฆ)=-๐ฆ, so ๐ข๐ฅ(๐ฅ,๐ฆ)=1๐ฃ๐ฅ(๐ฅ,๐ฆ)=0
๐ข๐ฆ(๐ฅ,๐ฆ)=0๐ฃ๐ฆ(๐ฅ,๐ฆ)=-1
Since ๐ข๐ฅ(๐ฅ,๐ฆ)โ ๐ฃ๐ฆ(๐ฅ,๐ฆ) (while ๐ข๐ฆ(๐ฅ,๐ฆ)=-๐ฃ๐ฅ(๐ฅ,๐ฆ) for all ๐ง, the function ๐ is not differentiable anywhere. Recall: If ๐ is differentiable at ๐ง0 then the Cauchy-Riemann equations hold at ๐ง0. However, if ๐ satisfies the Cauchy-Riemann equations at a point ๐ง0 then does this imply that ๐ is differentiable at ๐ง0? And what is the sufficient conditions for differentiability. By theorem. Let ๐=๐ข+๐๐ฃ be defined on a domain ๐ทโโ. Then ๐ is analytic in >๐ท if and only if ๐ข(๐ฅ,๐ฆ) and ๐ฃ(๐ฅ,๐ฆ) have continuous first partial derivatives on >๐ท that satisfy the Cauchy-Riemann equations. Example: ๐(๐ง)=๐๐ฅ ๐ข๐ฅ(๐ฅ,๐ฆ)=๐๐ฅ
Thus the Cauchy-Riemann equations are satisfied, and in addition, the functions ๐ข๐ฅ, ๐ข๐ฆ, ๐ฃ๐ฅ, ๐ฃ๐ฆ are continuous in โ. Therefore, the function ๐ is analytic in โ, thus entire. ยฉsideway ID: 190300030 Last Updated: 3/30/2019 Revision: 0 Latest Updated Links
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