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`The Cauchy-Riemann equationsβComplex FunctionβCauchy-Riemann equations`

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# The Cauchy-Riemann equations

## Complex Function

A complex function π can be written as π(π§)=π’(π₯,π¦)+ππ£(π₯,π¦) where π§=π₯+ππ¦ and π’, π£ are real-valued functions that depend on the two real variables π₯ and π¦.

Example:

`π(π§)=π§2=(π₯+ππ¦)2=π₯2-π¦2 π’(π₯,π¦) +πβ2π₯π¦ π£(π₯,π¦) `

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+π2=3+4π.

Alternatively, since π§=π₯+ππ¦, 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)+ππ£(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

• π is differentiable everywhere in β
• πβ²(π§)=2π§ for all π§ββ

For the function π’(π₯,π¦)

• If fix the variable π¦ at a certain value, then π’ only depends on π₯. For example, if only consider π¦=3, then π’(π₯,π¦)=π’(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 βπ’βπ₯(π₯,3)=𝑢π₯(π₯,3)=2π₯, and, more generally, for arbitrary (fixed) π¦, βπ’βπ₯(π₯,π¦)=π’π₯(π₯,π¦)=2π₯
• This is called the partial derivative of π’ with respect to π₯

Similarly, for the function π£(π₯,π¦)

• For example, π£(π₯,3)=2βπ₯β3=6π₯, and the derivative of this function with respect to π₯ is 6.
• Thus βπ£βπ₯(π₯,3)=π£π₯(π₯,3)=6
• More generally, for arbitrary (fixed) π¦, βπ£βπ₯(π₯,π¦)=π£π₯(π₯,π¦)=2π¦
• This is called the partial derivative of π£ with respect to π₯.

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

• Example: Let π₯ =2. Then π’(π₯,π¦)=π’(2,π¦)=4βπ¦2 and βπ’βπ¦(π₯,π¦)=π’π¦(2,π¦)=-2π¦
• More generally, βπ’βπ¦(π₯,π¦)=π’π¦(π₯,π¦)=-2π¦
• This is called the partial derivative of π’ with respect to π¦.
• Similarly, π£(2,π¦)=4π¦, and βπ£βπ¦(2,π¦)=π£π¦(2,π¦)=4. More generally, βπ£βπ¦(π₯,π¦)=π£π¦(π₯,π¦)=2π₯
• This is called the partial derivative of π£ 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:

`πβ²(π§)=2π§, π’π₯(π₯,π¦)=2π₯, π’π¦(π₯,π¦)=-2π¦, π£π₯(π₯,π¦)=2π¦, π£π¦(π₯,π¦)=2x`

Notice:

• π’π₯=π£π¦
• π’π¦=-π£π₯
• πβ²=π’π₯+ππ£π₯ =ππ₯ =-π(π’π¦+ππ£π¦) =-πππ¦

Another Example

```π(π§)=2π§3-4π§+1, where π§=π₯+ππ¦  =2(π₯+ππ¦)3-4(π₯+ππ¦)+1  =2(π₯3+3π₯2ππ¦+3π₯π2π¦2+π3π¦3)-4π₯-4ππ¦+1  =(2π₯3-6π₯π¦2-4π₯+1) π’(π₯,π¦) +π(6π₯2π¦-2π¦3-4π¦) π£(π₯,π¦) ```

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:

```πβ²(π§)=6π§2-4, where π§=π₯+ππ¦ =6(π₯+ππ¦)2-4  =(6π₯2-6π¦2-4)+12ππ₯π¦  =π’π₯(π₯,π¦)+ππ£π₯(π₯,π¦)=-π(π’π¦(π₯,π¦)+ππ£π¦(π₯,π¦))=ππ₯(π§)=-πππ¦(π§) ```

## Cauchy-Riemann equations

By Theorem. Suppose that π(π§)=π’(π₯,π¦)+ππ£(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,

```πβ²(π§0)=π’π₯(π₯0,π¦0)+ππ£π₯(π₯0,π¦0)=ππ₯(π§0)  =-π(π’π¦(π₯0,π¦0)+ππ£π¦(π₯0,π¦0))=-πππ¦(π§0)```

Method of Proof

For the difference quotient,

`π(π§0+β)-π(π§0)β`

whose limit as ββ0 must exist if π is differentiable at π§0

• Let β 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 π(π§)=π§=π₯-ππ¦, 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’+ππ£ be defined on a domain π·ββ. 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§)=ππ₯cosπ¦+πππ₯sinπ¦. The

```π’π₯(π₯,π¦)=ππ₯cosπ¦π£π₯(π₯,π¦)=ππ₯sinπ¦ π’π¦(π₯,π¦)=βππ₯sinπ¦π£π¦(π₯,π¦)=ππ₯cosπ¦ ```

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.

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

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