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The Cauchy-Riemann equations
 Complex Function
 Cauchy-Riemann equations

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The Cauchy-Riemann equations

Complex Function

A complex function 𝑓 can be written as 𝑓(𝑧)=𝑒(π‘₯,𝑦)+𝑖𝑣(π‘₯,𝑦) where 𝑧=π‘₯+𝑖𝑦 and 𝑒, 𝑣 are real-valued functions that depend on the two real variables π‘₯ and 𝑦.

Example:

𝑓(𝑧)=𝑧2=(π‘₯+𝑖𝑦)2=π‘₯2-𝑦2 𝑒(π‘₯,𝑦) +𝑖⋅2π‘₯𝑦 𝑣(π‘₯,𝑦)

Thus 𝑒(π‘₯,𝑦)=π‘₯2-𝑦2 and 𝑣(π‘₯,𝑦)=2π‘₯𝑦.

For example, if 𝑧=2+𝑖, then clearly 𝑓(𝑧)=(2+𝑖)2=4+4𝑖+𝑖2=3+4𝑖.

Alternatively, since 𝑧=π‘₯+𝑖𝑦, then π‘₯=2 and 𝑦=1, thus

𝑒(π‘₯,𝑦)=𝑒(2,1)=22-12=3 and 𝑣(π‘₯,𝑦)=𝑣(2,1)=2β‹…2β‹…1=4

thus

𝑓(2+𝑖)=𝑒(2,1)+𝑖𝑣(2,1)=3+4𝑖

Another example

𝑓(𝑧)=𝑧2=𝑒(π‘₯,𝑦)+𝑖=2𝑣(π‘₯,𝑦), 𝑒(π‘₯,𝑦)=π‘₯2-𝑦2, 𝑣(π‘₯,𝑦)=2π‘₯𝑦

Therefore

  • 𝑓 is differentiable everywhere in β„‚
  • 𝑓′(𝑧)=2𝑧 for all π‘§βˆˆβ„‚

For the function 𝑒(π‘₯,𝑦)

  • If fix the variable 𝑦 at a certain value, then 𝑒 only depends on π‘₯. For example, if only consider 𝑦=3, then 𝑒(π‘₯,𝑦)=𝑒(π‘₯,3)=π‘₯2-9
  • This function can now differentiate with respect to π‘₯ according to the rules of calculus and find that the derivative is 2π‘₯
  • That is βˆ‚π‘’βˆ‚π‘₯(π‘₯,3)=𝑢π‘₯(π‘₯,3)=2π‘₯, and, more generally, for arbitrary (fixed) 𝑦, βˆ‚π‘’βˆ‚π‘₯(π‘₯,𝑦)=𝑒π‘₯(π‘₯,𝑦)=2π‘₯
  • This is called the partial derivative of 𝑒 with respect to π‘₯

Similarly, for the function 𝑣(π‘₯,𝑦)

  • For example, 𝑣(π‘₯,3)=2β‹…π‘₯β‹…3=6π‘₯, and the derivative of this function with respect to π‘₯ is 6.
  • Thus βˆ‚π‘£βˆ‚π‘₯(π‘₯,3)=𝑣π‘₯(π‘₯,3)=6
  • More generally, for arbitrary (fixed) 𝑦, βˆ‚π‘£βˆ‚π‘₯(π‘₯,𝑦)=𝑣π‘₯(π‘₯,𝑦)=2𝑦
  • This is called the partial derivative of 𝑣 with respect to π‘₯.

Obviously, the same thing can be done by fixing π‘₯ and differentiating with respect to 𝑦.

  • Example: Let π‘₯ =2. Then 𝑒(π‘₯,𝑦)=𝑒(2,𝑦)=4βˆ’π‘¦2 and βˆ‚π‘’βˆ‚π‘¦(π‘₯,𝑦)=𝑒𝑦(2,𝑦)=-2𝑦
  • More generally, βˆ‚π‘’βˆ‚π‘¦(π‘₯,𝑦)=𝑒𝑦(π‘₯,𝑦)=-2𝑦
  • This is called the partial derivative of 𝑒 with respect to 𝑦.
  • Similarly, 𝑣(2,𝑦)=4𝑦, and βˆ‚π‘£βˆ‚π‘¦(2,𝑦)=𝑣𝑦(2,𝑦)=4. More generally, βˆ‚π‘£βˆ‚π‘¦(π‘₯,𝑦)=𝑣𝑦(π‘₯,𝑦)=2π‘₯
  • This is called the partial derivative of 𝑣 with respect to 𝑦.

And the result are

𝑓(𝑧)=𝑧2=𝑒(π‘₯,𝑦)+𝑖=2𝑣(π‘₯,𝑦), 𝑒(π‘₯,𝑦)=π‘₯2-𝑦2, 𝑣(π‘₯,𝑦)=2π‘₯𝑦

The derivatives:

𝑓′(𝑧)=2𝑧, 𝑒π‘₯(π‘₯,𝑦)=2π‘₯, 𝑒𝑦(π‘₯,𝑦)=-2𝑦, 𝑣π‘₯(π‘₯,𝑦)=2𝑦, 𝑣𝑦(π‘₯,𝑦)=2x

Notice:

  • 𝑒π‘₯=𝑣𝑦
  • 𝑒𝑦=-𝑣π‘₯
  • 𝑓′=𝑒π‘₯+𝑖𝑣π‘₯ =𝑓π‘₯ =-𝑖(𝑒𝑦+𝑖𝑣𝑦) =-𝑖𝑓𝑦

Another Example


𝑓(𝑧)=2𝑧3-4𝑧+1, where 𝑧=π‘₯+𝑖𝑦  =2(π‘₯+𝑖𝑦)3-4(π‘₯+𝑖𝑦)+1  =2(π‘₯3+3π‘₯2𝑖𝑦+3π‘₯𝑖2𝑦2+𝑖3𝑦3)-4π‘₯-4𝑖𝑦+1  =(2π‘₯3-6π‘₯𝑦2-4π‘₯+1) 𝑒(π‘₯,𝑦) +𝑖(6π‘₯2𝑦-2𝑦3-4𝑦) 𝑣(π‘₯,𝑦)

Then

𝑒π‘₯(π‘₯,𝑦)=6π‘₯2-6𝑦2-4𝑣π‘₯(π‘₯,𝑦)=12π‘₯𝑦 𝑒𝑦(π‘₯,𝑦)=-12π‘₯𝑦𝑣𝑦(π‘₯,𝑦)=6π‘₯2-6𝑦2-4

Thus, 𝑒π‘₯=𝑣𝑦 and 𝑒𝑦=-𝑣π‘₯

The derivatives:

𝑓′(𝑧)=6𝑧2-4, where 𝑧=π‘₯+𝑖𝑦 =6(π‘₯+𝑖𝑦)2-4  =(6π‘₯2-6𝑦2-4)+12𝑖π‘₯𝑦  =𝑒π‘₯(π‘₯,𝑦)+𝑖𝑣π‘₯(π‘₯,𝑦)=-𝑖(𝑒𝑦(π‘₯,𝑦)+𝑖𝑣𝑦(π‘₯,𝑦))=𝑓π‘₯(𝑧)=-𝑖𝑓𝑦(𝑧)

Cauchy-Riemann equations

By Theorem. Suppose that 𝑓(𝑧)=𝑒(π‘₯,𝑦)+𝑖𝑣(π‘₯,𝑦) is differentiable at a point 𝑧0. Then the partial derivatives 𝑒π‘₯, 𝑒𝑦, 𝑣π‘₯, 𝑣𝑦 exist at 𝑧0, and satisfy there:

𝑒π‘₯=𝑣𝑦 and 𝑒𝑦=-𝑣π‘₯

These are called the Cauchy-Riemann Equations. Also,

𝑓′(𝑧0)=𝑒π‘₯(π‘₯0,𝑦0)+𝑖𝑣π‘₯(π‘₯0,𝑦0)=𝑓π‘₯(𝑧0)  =-𝑖(𝑒𝑦(π‘₯0,𝑦0)+𝑖𝑣𝑦(π‘₯0,𝑦0))=-𝑖𝑓𝑦(𝑧0)

Method of Proof

For the difference quotient,

𝑓(𝑧0+β„Ž)-𝑓(𝑧0)β„Ž

whose limit as β„Žβ†’0 must exist if 𝑓 is differentiable at 𝑧0

  • Let β„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 𝑓(𝑧)=𝑧=π‘₯-𝑖𝑦, then 𝑒(π‘₯,𝑦)=π‘₯ and 𝑣(π‘₯,𝑦)=-𝑦, so

𝑒π‘₯(π‘₯,𝑦)=1𝑣π‘₯(π‘₯,𝑦)=0 𝑒𝑦(π‘₯,𝑦)=0𝑣𝑦(π‘₯,𝑦)=-1

Since 𝑒π‘₯(π‘₯,𝑦)≠𝑣𝑦(π‘₯,𝑦) (while 𝑒𝑦(π‘₯,𝑦)=-𝑣π‘₯(π‘₯,𝑦) for all 𝑧, the function 𝑓 is not differentiable anywhere.

Recall: If 𝑓 is differentiable at 𝑧0 then the Cauchy-Riemann equations hold at 𝑧0. However, if 𝑓 satisfies the Cauchy-Riemann equations at a point 𝑧0 then does this imply that 𝑓 is differentiable at 𝑧0? And what is the sufficient conditions for differentiability.

By theorem. Let 𝑓=𝑒+𝑖𝑣 be defined on a domain π·βŠ‚β„‚. Then 𝑓 is analytic in >𝐷 if and only if 𝑒(π‘₯,𝑦) and 𝑣(π‘₯,𝑦) have continuous first partial derivatives on >𝐷 that satisfy the Cauchy-Riemann equations.

Example: 𝑓(𝑧)=𝑒π‘₯cos𝑦+𝑖𝑒π‘₯sin𝑦. The

𝑒π‘₯(π‘₯,𝑦)=𝑒π‘₯cos𝑦𝑣π‘₯(π‘₯,𝑦)=𝑒π‘₯sin𝑦 𝑒𝑦(π‘₯,𝑦)=βˆ’π‘’π‘₯sin𝑦𝑣𝑦(π‘₯,𝑦)=𝑒π‘₯cos𝑦

Thus the Cauchy-Riemann equations are satisfied, and in addition, the functions 𝑒π‘₯, 𝑒𝑦, 𝑣π‘₯, 𝑣𝑦 are continuous in β„‚. Therefore, the function 𝑓 is analytic in β„‚, thus entire.


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


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