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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)=𝑢(𝑥,𝑦)+𝑖𝑣(𝑥,𝑦) where 𝑧=𝑥+𝑖𝑦 and 𝑢, 𝑣 are real-valued functions that depend on the two real variables 𝑥 and 𝑦.

Example:

`𝑓(𝑧)=𝑧2=(𝑥+𝑖𝑦)2=𝑥2-𝑦2 𝑢(𝑥,𝑦) +𝑖⋅2𝑥𝑦 𝑣(𝑥,𝑦) `

Thus 𝑢(𝑥,𝑦)=𝑥2-𝑦2 and 𝑣(𝑥,𝑦)=2𝑥𝑦.

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

Alternatively, since 𝑧=𝑥+𝑖𝑦, then 𝑥=2 and 𝑦=1, thus

`𝑢(𝑥,𝑦)=𝑢(2,1)=22-12=3 and 𝑣(𝑥,𝑦)=𝑣(2,1)=2⋅2⋅1=4`

thus

`𝑓(2+𝑖)=𝑢(2,1)+𝑖𝑣(2,1)=3+4𝑖`

Another example

`𝑓(𝑧)=𝑧2=𝑢(𝑥,𝑦)+𝑖=2𝑣(𝑥,𝑦), 𝑢(𝑥,𝑦)=𝑥2-𝑦2, 𝑣(𝑥,𝑦)=2𝑥𝑦`

Therefore

• 𝑓 is differentiable everywhere in
• 𝑓′(𝑧)=2𝑧 for all 𝑧∈ℂ

For the function 𝑢(𝑥,𝑦)

• If fix the variable 𝑦 at a certain value, then 𝑢 only depends on 𝑥. For example, if only consider 𝑦=3, then 𝑢(𝑥,𝑦)=𝑢(𝑥,3)=𝑥2-9
• This function can now differentiate with respect to 𝑥 according to the rules of calculus and find that the derivative is 2𝑥
• That is ∂𝑢∂𝑥(𝑥,3)=𝑢𝑥(𝑥,3)=2𝑥, and, more generally, for arbitrary (fixed) 𝑦, ∂𝑢∂𝑥(𝑥,𝑦)=𝑢𝑥(𝑥,𝑦)=2𝑥
• This is called the partial derivative of 𝑢 with respect to 𝑥

Similarly, for the function 𝑣(𝑥,𝑦)

• For example, 𝑣(𝑥,3)=2⋅𝑥⋅3=6𝑥, and the derivative of this function with respect to 𝑥 is 6.
• Thus ∂𝑣∂𝑥(𝑥,3)=𝑣𝑥(𝑥,3)=6
• More generally, for arbitrary (fixed) 𝑦, ∂𝑣∂𝑥(𝑥,𝑦)=𝑣𝑥(𝑥,𝑦)=2𝑦
• This is called the partial derivative of 𝑣 with respect to 𝑥.

Obviously, the same thing can be done by fixing 𝑥 and differentiating with respect to 𝑦.

• Example: Let 𝑥 =2. Then 𝑢(𝑥,𝑦)=𝑢(2,𝑦)=4−𝑦2 and ∂𝑢∂𝑦(𝑥,𝑦)=𝑢𝑦(2,𝑦)=-2𝑦
• More generally, ∂𝑢∂𝑦(𝑥,𝑦)=𝑢𝑦(𝑥,𝑦)=-2𝑦
• This is called the partial derivative of 𝑢 with respect to 𝑦.
• Similarly, 𝑣(2,𝑦)=4𝑦, and ∂𝑣∂𝑦(2,𝑦)=𝑣𝑦(2,𝑦)=4. More generally, ∂𝑣∂𝑦(𝑥,𝑦)=𝑣𝑦(𝑥,𝑦)=2𝑥
• This is called the partial derivative of 𝑣 with respect to 𝑦.

And the result are

`𝑓(𝑧)=𝑧2=𝑢(𝑥,𝑦)+𝑖=2𝑣(𝑥,𝑦), 𝑢(𝑥,𝑦)=𝑥2-𝑦2, 𝑣(𝑥,𝑦)=2𝑥𝑦`

The derivatives:

`𝑓′(𝑧)=2𝑧, 𝑢𝑥(𝑥,𝑦)=2𝑥, 𝑢𝑦(𝑥,𝑦)=-2𝑦, 𝑣𝑥(𝑥,𝑦)=2𝑦, 𝑣𝑦(𝑥,𝑦)=2x`

Notice:

• 𝑢𝑥=𝑣𝑦
• 𝑢𝑦=-𝑣𝑥
• 𝑓′=𝑢𝑥+𝑖𝑣𝑥 =𝑓𝑥 =-𝑖(𝑢𝑦+𝑖𝑣𝑦) =-𝑖𝑓𝑦

Another Example

```𝑓(𝑧)=2𝑧3-4𝑧+1, where 𝑧=𝑥+𝑖𝑦  =2(𝑥+𝑖𝑦)3-4(𝑥+𝑖𝑦)+1  =2(𝑥3+3𝑥2𝑖𝑦+3𝑥𝑖2𝑦2+𝑖3𝑦3)-4𝑥-4𝑖𝑦+1  =(2𝑥3-6𝑥𝑦2-4𝑥+1) 𝑢(𝑥,𝑦) +𝑖(6𝑥2𝑦-2𝑦3-4𝑦) 𝑣(𝑥,𝑦) ```

Then

```𝑢𝑥(𝑥,𝑦)=6𝑥2-6𝑦2-4𝑣𝑥(𝑥,𝑦)=12𝑥𝑦 𝑢𝑦(𝑥,𝑦)=-12𝑥𝑦𝑣𝑦(𝑥,𝑦)=6𝑥2-6𝑦2-4 ```

Thus, 𝑢𝑥=𝑣𝑦 and 𝑢𝑦=-𝑣𝑥

The derivatives:

```𝑓′(𝑧)=6𝑧2-4, where 𝑧=𝑥+𝑖𝑦 =6(𝑥+𝑖𝑦)2-4  =(6𝑥2-6𝑦2-4)+12𝑖𝑥𝑦  =𝑢𝑥(𝑥,𝑦)+𝑖𝑣𝑥(𝑥,𝑦)=-𝑖(𝑢𝑦(𝑥,𝑦)+𝑖𝑣𝑦(𝑥,𝑦))=𝑓𝑥(𝑧)=-𝑖𝑓𝑦(𝑧) ```

## Cauchy-Riemann equations

By Theorem. Suppose that 𝑓(𝑧)=𝑢(𝑥,𝑦)+𝑖𝑣(𝑥,𝑦) is differentiable at a point 𝑧0. Then the partial derivatives 𝑢𝑥, 𝑢𝑦, 𝑣𝑥, 𝑣𝑦 exist at 𝑧0, and satisfy there:

`𝑢𝑥=𝑣𝑦 and 𝑢𝑦=-𝑣𝑥`

These are called the Cauchy-Riemann Equations. Also,

```𝑓′(𝑧0)=𝑢𝑥(𝑥0,𝑦0)+𝑖𝑣𝑥(𝑥0,𝑦0)=𝑓𝑥(𝑧0)  =-𝑖(𝑢𝑦(𝑥0,𝑦0)+𝑖𝑣𝑦(𝑥0,𝑦0))=-𝑖𝑓𝑦(𝑧0)```

Method of Proof

For the difference quotient,

`𝑓(𝑧0+ℎ)-𝑓(𝑧0)ℎ`

whose limit as ℎ→0 must exist if 𝑓 is differentiable at 𝑧0

• Let ℎ 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=𝑥-𝑖𝑦, then 𝑢(𝑥,𝑦)=𝑥 and 𝑣(𝑥,𝑦)=-𝑦, so

```𝑢𝑥(𝑥,𝑦)=1𝑣𝑥(𝑥,𝑦)=0 𝑢𝑦(𝑥,𝑦)=0𝑣𝑦(𝑥,𝑦)=-1 ```

Since 𝑢𝑥(𝑥,𝑦)≠𝑣𝑦(𝑥,𝑦) (while 𝑢𝑦(𝑥,𝑦)=-𝑣𝑥(𝑥,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=𝑢+𝑖𝑣 be defined on a domain 𝐷⊂ℂ. Then 𝑓 is analytic in >𝐷 if and only if 𝑢(𝑥,𝑦) and 𝑣(𝑥,𝑦) have continuous first partial derivatives on >𝐷 that satisfy the Cauchy-Riemann equations.

Example: 𝑓(𝑧)=𝑒𝑥cos𝑦+𝑖𝑒𝑥sin𝑦. The

```𝑢𝑥(𝑥,𝑦)=𝑒𝑥cos𝑦𝑣𝑥(𝑥,𝑦)=𝑒𝑥sin𝑦 𝑢𝑦(𝑥,𝑦)=−𝑒𝑥sin𝑦𝑣𝑦(𝑥,𝑦)=𝑒𝑥cos𝑦 ```

Thus the Cauchy-Riemann equations are satisfied, and in addition, the functions 𝑢𝑥, 𝑢𝑦, 𝑣𝑥, 𝑣𝑦 are continuous in . Therefore, the function 𝑓 is analytic in ℂ, thus entire.

ID: 190300030 Last Updated: 30/3/2019 Revision: 0 Home 5

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