Complex AnalysisComplex NumberTopologyFunctionSequences and LimitsIteration of FunctionComplex Derivative Draft for Information Only
ContentThe Cauchy-Riemann equations
source/reference: The Cauchy-Riemann equationsComplex FunctionA complex function π can be written as π(π§)=π’(π₯,π¦)+ππ£(π₯,π¦) where π§=π₯+ππ¦ and π’, π£ are real-valued functions that depend on the two real variables π₯ and π¦. Example: π(π§)=π§2=(π₯+ππ¦)2=
Thus π’(π₯,π¦)=π₯2-π¦2 and π£(π₯,π¦)=2π₯π¦. For example, if π§=2+π, then clearly π(π§)=(2+π)2=4+4π+π2=3+4π. Alternatively, since π§=π₯+ππ¦, then π₯=2 and π¦=1, thus π’(π₯,π¦)=π’(2,1)=22-12=3 and π£(π₯,π¦)=π£(2,1)=2β
2β
1=4
thus π(2+π)=π’(2,1)+ππ£(2,1)=3+4π
Another example π(π§)=π§2=π’(π₯,π¦)+π=2π£(π₯,π¦), π’(π₯,π¦)=π₯2-π¦2, π£(π₯,π¦)=2π₯π¦
Therefore
For the function π’(π₯,π¦)
Similarly, for the function π£(π₯,π¦)
Obviously, the same thing can be done by fixing π₯ and differentiating with respect to π¦.
And the result are π(π§)=π§2=π’(π₯,π¦)+π=2π£(π₯,π¦), π’(π₯,π¦)=π₯2-π¦2, π£(π₯,π¦)=2π₯π¦
The derivatives: πβ²(π§)=2π§, π’π₯(π₯,π¦)=2π₯, π’π¦(π₯,π¦)=-2π¦, π£π₯(π₯,π¦)=2π¦, π£π¦(π₯,π¦)=2x
Notice:
Another Example π(π§)=2π§3-4π§+1, where π§=π₯+ππ¦
=2(π₯+ππ¦)3-4(π₯+ππ¦)+1
=2(π₯3+3π₯2ππ¦+3π₯π2π¦2+π3π¦3)-4π₯-4ππ¦+1
=
Then π’π₯(π₯,π¦)=6π₯2-6π¦2-4π£π₯(π₯,π¦)=12π₯π¦
π’π¦(π₯,π¦)=-12π₯π¦π£π¦(π₯,π¦)=6π₯2-6π¦2-4
Thus, π’π₯=π£π¦ and π’π¦=-π£π₯ The derivatives: πβ²(π§)=6π§2-4, where π§=π₯+ππ¦ =6(π₯+ππ¦)2-4
=(6π₯2-6π¦2-4)+12ππ₯π¦
=π’π₯(π₯,π¦)+ππ£π₯(π₯,π¦)=-π(π’π¦(π₯,π¦)+ππ£π¦(π₯,π¦))=ππ₯(π§)=-πππ¦(π§)
Cauchy-Riemann equationsBy Theorem. Suppose that π(π§)=π’(π₯,π¦)+ππ£(π₯,π¦) is differentiable at a point π§0. Then the partial derivatives π’π₯, π’π¦, π£π₯, π£π¦ exist at π§0, and satisfy there: π’π₯=π£π¦ and π’π¦=-π£π₯
These are called the Cauchy-Riemann Equations. Also, πβ²(π§0)=π’π₯(π₯0,π¦0)+ππ£π₯(π₯0,π¦0)=ππ₯(π§0)
=-π(π’π¦(π₯0,π¦0)+ππ£π¦(π₯0,π¦0))=-πππ¦(π§0)
Method of Proof For the difference quotient,
whose limit as ββ0 must exist if π is differentiable at π§0
Another example Let π(π§)=π§=π₯-ππ¦, then π’(π₯,π¦)=π₯ and π£(π₯,π¦)=-π¦, so π’π₯(π₯,π¦)=1π£π₯(π₯,π¦)=0
π’π¦(π₯,π¦)=0π£π¦(π₯,π¦)=-1
Since π’π₯(π₯,π¦)β π£π¦(π₯,π¦) (while π’π¦(π₯,π¦)=-π£π₯(π₯,π¦) for all π§, the function π is not differentiable anywhere. Recall: If π is differentiable at π§0 then the Cauchy-Riemann equations hold at π§0. However, if π satisfies the Cauchy-Riemann equations at a point π§0 then does this imply that π is differentiable at π§0? And what is the sufficient conditions for differentiability. By theorem. Let π=π’+ππ£ be defined on a domain π·ββ. Then π is analytic in >π· if and only if π’(π₯,π¦) and π£(π₯,π¦) have continuous first partial derivatives on >π· that satisfy the Cauchy-Riemann equations. Example: π(π§)=ππ₯ π’π₯(π₯,π¦)=ππ₯
Thus the Cauchy-Riemann equations are satisfied, and in addition, the functions π’π₯, π’π¦, π£π₯, π£π¦ are continuous in β. Therefore, the function π is analytic in β, thus entire. Β©sideway ID: 190300030 Last Updated: 3/30/2019 Revision: 0 Latest Updated Links
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