Cauchy-Riemann-Differential equation

Now, if we decompose a function ${\displaystyle f:U\rightarrow \mathbb {C} }$ with ${\displaystyle f\left(x+iy\right)=u\left(x,y\right)+i,v\left(x,y\right)}$ into its real and imaginary parts with real functions ${\displaystyle u:U_{R}\rightarrow \mathbb {R} }$, ${\displaystyle v:U_{R}\rightarrow \mathbb {R} }$ where ${\displaystyle U_{R}\subset \mathbb {R} ^{2}}$ and ${\displaystyle U={x+iy\in \mathbb {C} \ |\ (x,y)\in U_{R}}}$, then the total derivative of the function ${\displaystyle f_{R}:U_{R}\rightarrow \mathbb {R} ^{2},(x,y)\mapsto {\begin{pmatrix}u\left(x,y\right)\ v\left(x,y\right)\end{pmatrix}}}$ has the following Jacobian matrix as its representation: ${\displaystyle {\begin{pmatrix}{\frac {\partial u}{\partial x}}&{\frac {\partial u}{\partial y}}\\{\frac {\partial v}{\partial x}}&{\frac {\partial v}{\partial y}}\end{pmatrix}}.}$

The evaluation of the Jacobian matrix at a point ${\displaystyle (x_{o},y_{o})\in \mathbb {R} ^{2}}$ gives the total derivative at the point ${\displaystyle x_{o}+iy_{o}\in \mathbb {C} }$ :${\displaystyle {\begin{pmatrix}{\frac {\partial u}{\partial x}}(x_{o},y_{o})&{\frac {\partial u}{\partial y}}(x_{o},y_{o})\ {\frac {\partial v}{\partial x}}(x_{o},y_{o})&{\frac {\partial v}{\partial y}}(x_{o},y_{o})\end{pmatrix}}}$

A function ${\displaystyle f}$ is complex differentiable at ${\displaystyle z_{o}:=x_{o}+iy_{o}}$ if and only if it is real differentiable and the Cauchy-Riemann differential equations hold for ${\displaystyle u,v}$ with ${\displaystyle u:U_{R}\rightarrow \mathbb {R} }$, ${\displaystyle v:U_{R}\rightarrow \mathbb {R} }$ where ${\displaystyle U_{R}\subset \mathbb {R} ^{2}}$: :${\displaystyle {\frac {\partial u}{\partial x}}(x_{o},y_{o})={\frac {\partial v}{\partial y}}(x_{o},y_{o})}$ :${\displaystyle \displaystyle {\frac {\partial u}{\partial y}}(x_{o},y_{o})=-{\frac {\partial v}{\partial x}}(x_{o},y_{o})}$ are satisfied.

If the following limit exists for ${\displaystyle f:G\rightarrow \mathbb {C} }$ at ${\displaystyle z_{o}\in G}$ with ${\displaystyle G\subset \mathbb {C} }$ open: :${\displaystyle f'(z_{o})=\lim _{z\rightarrow z_{o}}{\frac {f(z)-f(z_{o})}{z-z_{o}}}}$, then for any sequences ${\displaystyle (z_{n}){n\in \mathbb {N} }}$ in the domain ${\displaystyle G\subset \mathbb {C} }$ with ${\displaystyle \lim {n\rightarrow \infty }z_{n}=z_{o}}$, we also have: :${\displaystyle f'(z_{o})=\lim _{n\rightarrow \infty }{\frac {f(z_{n})-f(z_{o})}{z_{n}-z_{o}}}}$

Now consider only the sequences for the two following limit processes with ${\displaystyle h\in \mathbb {R} }$: :${\displaystyle f'(z_{o})=\lim _{h\rightarrow 0}{\frac {f(z_{o}+h)-f(z_{o})}{(z_{o}+h)-z_{o}}}=\lim _{h\rightarrow 0}{\frac {f(z_{o}+h)-f(z_{o})}{h}}}$, :${\displaystyle f'(z_{o})=\lim _{ih\rightarrow 0}{\frac {f(z_{o}+ih)-f(z_{o})}{(z_{o}+ih)-z_{o}}}=\lim _{ih\rightarrow 0}{\frac {f(z_{o}+ih)-f(z_{o})}{ih}}}$,

By inserting the component functions for the real and imaginary parts ${\displaystyle u,v}$, we get with ${\displaystyle h\in \mathbb {R} }$: :${\displaystyle f'(z_{o})=\lim _{h\rightarrow 0}{\frac {f(z_{o}+h)-f(z_{o})}{h}}=}$ ::${\displaystyle =\lim _{h\rightarrow 0}{\frac {u(x_{o}+h,y_{o})-u(x_{o},y_{o})}{h}}+i\lim _{h\rightarrow 0}{\frac {v(x_{o}+h,y_{o})-v(x_{o},y_{o})}{h}}}$ ::${\displaystyle ={\frac {\partial u}{\partial x}}(x_{o},y_{o})+i{\frac {\partial v}{\partial x}}(x_{o},y_{o})}$

Applying this to the second equation, we get with ${\displaystyle h\in \mathbb {R} }$: :${\displaystyle f'(z_{o})=\lim _{ih\rightarrow 0}{\frac {f(z_{o}+ih)-f(z_{o})}{ih}}}$ ::${\displaystyle =\lim _{h\rightarrow 0}{\frac {u(x_{o},y_{o}+h)-u(x_{o},y_{o})}{ih}}+i\lim _{h\rightarrow 0}{\frac {v(x_{o},y_{o}+h)-v(x_{o},y_{o})}{ih}}}$ ::${\displaystyle =-i\lim _{h\rightarrow 0}{\frac {u(x_{o},y_{o}+h)-u(x_{o},y_{o})}{h}}+\lim _{h\rightarrow 0}{\frac {v(x_{o},y_{o}+h)-v(x_{o},y_{o})}{h}}}$,

${\displaystyle =-i{\frac {\partial u}{\partial y}}(x_{o},y_{o})+{\frac {\partial v}{\partial y}}(x_{o},y_{o})}$

In the first summand, the fraction is extended by ${\displaystyle i}$ , and in the second summand ${\displaystyle i}$, the is canceled so that the denominator becomes real-valued and ${\displaystyle h}$ corresponds.

By equating the terms from (3) and (4) and comparing the real and imaginary parts, we obtain the Cauchy-Riemann differential equations.

• Real part: ${\displaystyle {\frac {\partial u}{\partial x}}(x_{o},y_{o})={\frac {\partial v}{\partial y}}(x_{o},y_{o})}$
• Imaginary part: ${\displaystyle \displaystyle {\frac {\partial u}{\partial y}}(x_{o},y_{o})=-{\frac {\partial v}{\partial x}}(x_{o},y_{o})}$

The partial derivatives in ${\displaystyle \mathbb {R} ^{2}}$ of the Cauchy-Riemann differential equations can also be expressed in ${\displaystyle \mathbb {C} }$ with ${\displaystyle f:={\mathfrak {Re}}(f)+i{\mathfrak {Im}}(f)}$, ${\displaystyle {\mathfrak {Re}}(f):\mathbb {C} \rightarrow \mathbb {R} }$, ${\displaystyle {\mathfrak {Im}}(f):\mathbb {C} \rightarrow \mathbb {R} }$, and ${\displaystyle h\in \mathbb {R} }$.

${\displaystyle {\frac {\partial f}{\partial x}}(z_{o})=\lim _{h\rightarrow 0}{\frac {f(z_{o}+h)-f(z_{o})}{h}}\in \mathbb {C} }$,

${\displaystyle {\frac {\partial {\mathfrak {Re}}(f)}{\partial x}}(z_{o})=\lim _{h\rightarrow 0}{\frac {{\mathfrak {Re}}(f)(z_{o}+h)-{\mathfrak {Re}}(f)(z_{o})}{h}}\in \mathbb {R} }$,

${\displaystyle {\frac {\partial {\mathfrak {Im}}(f)}{\partial x}}(z_{o})=\lim _{h\rightarrow 0}{\frac {{\mathfrak {Im}}(f)(z_{o}+h)-{\mathfrak {Im}}(f)(z_{o})}{h}}\in \mathbb {R} }$.

The partial derivatives in ${\displaystyle \mathbb {R} ^{2}}$ of the Cauchy-Riemann differential equations can also be expressed in ${\displaystyle \mathbb {C} }$ with ${\displaystyle f:={\mathfrak {Re}}(f)+i{\mathfrak {Im}}(f)}$, ${\displaystyle {\mathfrak {Re}}(f):\mathbb {C} \rightarrow \mathbb {R} }$, ${\displaystyle {\mathfrak {Im}}(f):\mathbb {C} \rightarrow \mathbb {R} }$, and ${\displaystyle h\in \mathbb {R} }$.

${\displaystyle {\frac {\partial f}{\partial y}}(z_{o})=\lim _{h\rightarrow 0}{\frac {f(z_{o}+ih)-f(z_{o})}{h}}\in \mathbb {C} }$,

${\displaystyle {\frac {\partial {\mathfrak {Re}}(f)}{\partial y}}(z_{o})=\lim _{h\rightarrow 0}{\frac {{\mathfrak {Re}}(f)(z_{o}+ih)-{\mathfrak {Re}}(f)(z_{o})}{h}}\in \mathbb {R} }$,

${\displaystyle {\frac {\partial {\mathfrak {Im}}(f)}{\partial y}}(z_{o})=\lim _{h\rightarrow 0}{\frac {{\mathfrak {Im}}(f)(z_{o}+ih)-{\mathfrak {Im}}(f)(z_{o})}{h}}\in \mathbb {R} }$.

The partial derivatives of the Cauchy-Riemann differential equations can also be expressed in ${\displaystyle \mathbb {C} }$ with ${\displaystyle f:=f_{x}+if_{y}}$, ${\displaystyle f_{x}:={\mathfrak {Re}}(f)}$, ${\displaystyle f_{y}:={\mathfrak {Im}}(f)}$: Real part: ${\displaystyle {\frac {\partial {\mathfrak {Re}}(f)}{\partial x}}(z_{o})={\frac {\partial {\mathfrak {Im}}(f)}{\partial y}}(z_{o})}$ Imaginary part: ${\displaystyle \displaystyle {\frac {\partial {\mathfrak {Re}}(f)}{\partial y}}(z_{o})=-{\frac {\partial {\mathfrak {Im}}(f)}{\partial x}}(z_{o})}$

In this case, the derivative of ${\textstyle f}$ at the point ${\textstyle z\in \mathbb {C} }$ can be represented in two ways using the component functions ${\textstyle u}$ and ${\textstyle v}$: $${\displaystyle f'\left(z\right)={\dfrac {\partial u}{\partial x}}\left(x,y\right)-i{\dfrac {\partial u}{\partial y}}\left(x,y\right)={\dfrac {\partial v}{\partial y}}\left(x,y\right)+i{\dfrac {\partial v}{\partial x}}\left(x,y\right)}$$ The proof of the Cauchy-Riemann differential equations uses a comparison of the real and imaginary parts to derive the above equations.

In the first step, let ${\displaystyle h\in \mathbb {C} }$ converge to 0 in the direction of the real part. To achieve this, choose ${\textstyle h:=h_{1}+i\cdot 0}$ with ${\textstyle h_{1}\in \mathbb {R} }$. The decomposition of the function ${\textstyle f=u+iv}$ into its real part ${\displaystyle u}$ and imaginary part ${\displaystyle v}$ then yields (DG1).

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

Similarly, the partial derivative for the imaginary part can be considered with ${\textstyle h:=0+i\cdot h_{2}}$ and ${\textstyle h_{2}\in \mathbb {R} }$. This yields equation (DG2).

${\textstyle {\begin{array}{rcl}f'\left(z\right)&=&\lim \limits _{h\to 0}{\dfrac {f\left(z+h\right)-f\left(z\right)}{h}}\\&=&\lim \limits _{h_{2}\to 0}{\dfrac {u\left(x,y+h_{2}\right)+iv\left(x,y+h_{2}\right)-u\left(x,y\right)-iv\left(x,y\right)}{i\cdot h_{2}}}\\&=&\lim \limits _{l\to 0}{\dfrac {1}{i}}{\dfrac {u\left(x,y+h_{2}\right)-u\left(x,y\right)}{h_{2}}}+{\dfrac {v\left(x,y+h_{2}\right)-v\left(x,y\right)}{h_{2}}}\\&=&{\dfrac {\partial v}{\partial y}}\left(x,y\right)-i{\dfrac {\partial u}{\partial y}}\left(x,y\right)\end{array}}}$

By equating the two derivatives, one can compare the real and imaginary parts of the two derivatives (DG1) and (DG2): $${\displaystyle f'\left(z\right)={\dfrac {\partial u}{\partial x}}\left(x,y\right)+i{\dfrac {\partial v}{\partial x}}\left(x,y\right)={\dfrac {\partial v}{\partial y}}\left(x,y\right)-i{\dfrac {\partial u}{\partial y}}\left(x,y\right)}$$

Two complex numbers are equal if and only if their real and imaginary parts are equal. This results in the Cauchy-Riemann differential equations. The two representation formulas follow from the above equation and the Cauchy-Riemann equations.

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• Date: 12/26/2024