Cauchy-Riemann Equations

## Theorem

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Let  G ⊆ C {\displaystyle G\subseteq \mathbb {C} } be an open subset. Let the function  f = u + i v {\displaystyle f=u+iv} be differentiable at a point  z = x + i y ∈ G {\displaystyle z=x+iy\in G} . Then all partial derivatives of  u {\displaystyle u} and  v {\displaystyle v} exist at  ( x , y ) {\displaystyle \left(x,y\right)} and the following Cauchy-Riemann equations hold:

∂ u ∂ x ( x , y ) = ∂ v ∂ y ( x , y ) ∂ u ∂ y ( x , y ) = − ∂ v ∂ x ( x , y ) {\displaystyle {\dfrac {\partial u}{\partial x}}\left(x,y\right)={\dfrac {\partial v}{\partial y}}\left(x,y\right)\quad \quad {\dfrac {\partial u}{\partial y}}\left(x,y\right)=-{\dfrac {\partial v}{\partial x}}\left(x,y\right)}

In this case, the derivative of  f {\displaystyle f} at  z {\displaystyle z} can be represented by the formula

f ′ ( z ) = ∂ u ∂ x ( x , y ) − i ∂ u ∂ y ( x , y ) = ∂ v ∂ y ( x , y ) + i ∂ v ∂ x ( x , y ) {\displaystyle f'\left(z\right)={\dfrac {\partial u}{\partial x}}\left(x,y\right)-i{\dfrac {\partial u}{\partial y}}\left(x,y\right)={\dfrac {\partial v}{\partial y}}\left(x,y\right)+i{\dfrac {\partial v}{\partial x}}\left(x,y\right)}

## Proof

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The proof can be decomposed into 3 main steps:

* calculate the partial derivative for the real part,
* calculate the partial derivative for the imaginary part,
* due to property of  f {\displaystyle f} being complex differentiable both derivatives yield the same complex value. This leads to Cauchy-Riemann equations.

### Proof - Step 1

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Let  h := k + i 0 ( k ∈ R ) {\displaystyle h:=k+i0\left(k\in \mathbb {R} \right)} . Then

f ′ ( z ) = lim h → 0 f ( z + h ) − f ( z ) h = lim k → 0 u ( x + k , y ) + i v ( x + k , y ) − u ( x , y ) − i v ( x , y ) k = lim k → 0 u ( x + k , y ) − u ( x , y ) k + i v ( x + k , y ) − v ( x , y ) k = ∂ u ∂ x ( x , y ) + i ∂ v ∂ x ( x , y ) {\displaystyle {\begin{array}{rcl}f'\left(z\right)&=&\lim \limits _{h\to 0}{\dfrac {f\left(z+h\right)-f\left(z\right)}{h}}\\\&=&\lim \limits _{k\to 0}{\dfrac {u\left(x+k,y\right)+iv\left(x+k,y\right)-u\left(x,y\right)-iv\left(x,y\right)}{k}}\\\&=&\lim \limits _{k\to 0}{\dfrac {u\left(x+k,y\right)-u\left(x,y\right)}{k}}+i{\dfrac {v\left(x+k,y\right)-v\left(x,y\right)}{k}}\\\&=&{\dfrac {\partial u}{\partial x}}\left(x,y\right)+i{\dfrac {\partial v}{\partial x}}\left(x,y\right)\end{array}}}

### Proof - Step 2

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Let  h := 0 + i l ( l ∈ R ) {\displaystyle h:=0+il\left(l\in \mathbb {R} \right)} . Then

f ′ ( z ) = lim h → 0 f ( z + h ) − f ( z ) h = lim l → 0 u ( x , y + l ) + i v ( x , y + l ) − u ( x , y ) − i v ( x , y ) i l = lim l → 0 1 i u ( x , y + l ) − u ( x , y ) l + v ( x , y + l ) − v ( x , y ) l = ∂ v ∂ y ( x , y ) − i ∂ u ∂ y ( x , y ) {\displaystyle {\begin{array}{rcl}f'\left(z\right)&=&\lim \limits _{h\to 0}{\dfrac {f\left(z+h\right)-f\left(z\right)}{h}}\\\&=&\lim \limits _{l\to 0}{\dfrac {u\left(x,y+l\right)+iv\left(x,y+l\right)-u\left(x,y\right)-iv\left(x,y\right)}{il}}\\\&=&\lim \limits _{l\to 0}{\dfrac {1}{i}}{\dfrac {u\left(x,y+l\right)-u\left(x,y\right)}{l}}+{\dfrac {v\left(x,y+l\right)-v\left(x,y\right)}{l}}\\\&=&{\dfrac {\partial v}{\partial y}}\left(x,y\right)-i{\dfrac {\partial u}{\partial y}}\left(x,y\right)\end{array}}}

### Proof - Step 3

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Both partial derivatives must provide a same complex value due to the fact that  f {\displaystyle f} is complex differentiable:  f ′ ( z ) = ∂ u ∂ x ( x , y ) + i ∂ v ∂ x ( x , y ) = ∂ v ∂ y ( x , y ) − i ∂ u ∂ y ( x , y ) {\displaystyle f'\left(z\right)={\dfrac {\partial u}{\partial x}}\left(x,y\right)+i{\dfrac {\partial v}{\partial x}}\left(x,y\right)={\dfrac {\partial v}{\partial y}}\left(x,y\right)-i{\dfrac {\partial u}{\partial y}}\left(x,y\right)}

### Proof - Step 4

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Equating the real and imaginary parts, we get the Cauchy-Riemann equations. The representation formula follows from the above line and the Cauchy-Riemann equations.

## See also

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* Complex Analysis

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