Cauchy-Riemann equation

 

Def 1

For $f(z) = u(x,y) + i v(x, y)$ and $z_{0} = x_{0} + iy_{0}$,

$$\frac{\partial f}{\partial x}(x_{0}, y_{0}) = -i \frac{\partial f}{\partial y}(x_{0}, y_{0})$$

$$( \; \Leftrightarrow  \frac{\partial u}{\partial x}(x_{0}, y_{0}) = \frac{\partial v}{\partial y}(x_{0}, y_{0}), \; \frac{\partial v}{\partial x}(x_{0}, y_{0}) = - \frac{\partial u}{\partial y}(x_{0}, y_{0}) \; )$$

called Cauchy-Riemann equation.

 

# also called C-R equation

 

 

 

Thm 1

$f(z) = u(x, y) + iv(x, y)$ is differentiable at $z_{0} = x_{0} + iy_{0}$.

$\Rightarrow$  The partial derivatives of $u, v$ exist at $z_{0}$

 

  $f'(z_{0}) = \displaystyle \lim_{z \to z_{0}} \frac{f(z) - f(z_{0})}{z - z_{0}}$
                $= \displaystyle \lim_{x + iy_{0} \to z_{0}} \frac{f(x + iy_{0}) - f(z_{0})}{x + iy_{0} - z_{0}}$
                $= \displaystyle \lim_{x \to x_{0}} \frac{f(x + iy_{0}) - f(x_{0} + iy_{0})}{x - x_{0}}$
                $= \displaystyle \lim_{x \to x_{0}} \frac{u(x, y_{0}) - u(x_{0}, y_{0})}{x - x_{0}} + i \frac{v(x, y_{0}) - v(x_{0}, y_{0})}{x - x_{0}} $

 

  By Prop 1,

  $\displaystyle \lim_{x \to x_{0}} \frac{u(x, y_{0}) - u(x_{0}, y_{0})}{x - x_{0}} = \mathrm{Re} \; f'(z_{0})$

  $\displaystyle \lim_{x \to x_{0}} \frac{v(x, y_{0}) - v(x_{0}, y_{0})}{x - x_{0}} = \mathrm{Im} \; f'(z_{0})$

 

  Also

  $f'(z_{0}) = \displaystyle \lim_{z \to z_{0}} \frac{f(z) - f(z_{0})}{z - z_{0}}$
                $= \displaystyle \lim_{x_{0} + iy \to z_{0}} \frac{f(x_{0} + iy) - f(z_{0})}{x_{0} + iy - z_{0}}$
                $= \displaystyle \lim_{y \to y_{0}} \frac{f(x_{0} + iy) - f(x_{0} + iy_{0})}{i(y - y_{0})}$
                $= \displaystyle \lim_{y \to y_{0}} -i\frac{u(x_{0}, y) - u(x_{0}, y_{0})}{y - y_{0}} +  \frac{v(x_{0}, y) - v(x_{0}, y_{0})}{y - y_{0}} $

 

  By Prop 1,

  $\displaystyle \lim_{y \to y_{0}} \frac{u(x_{0}, y) - u(x_{0}, y_{0})}{y - y_{0}} = -\mathrm{Im} \; f'(z_{0})$

  $\displaystyle \lim_{y \to y_{0}} \frac{v(x_{0}, y) - v(x_{0}, y_{0})}{y - y_{0}} = \mathrm{Re} \; f'(z_{0})$

 

  $\therefore$  The partial derivatives of $u, v$ exist at $z_{0}$

 

 

 

Thm 2

$f(z) = u(x,y) + i v(x, y)$ is differentiable at $z_{0} = x_{0} + iy_{0}$

$\Rightarrow$  C-R equation is satisfied at $z_{0}$ 

 

  $f'(z_{0}) = \displaystyle \lim_{z \to z_{0}} \frac{f(z) - f(z_{0})}{z - z_{0}}$
                $= \displaystyle \lim_{x + iy_{0} \to z_{0}} \frac{f(x + iy_{0}) - f(z_{0})}{x + iy_{0} - z_{0}}$
                $= \displaystyle \lim_{x \to x_{0}} \frac{f(x + iy_{0}) - f(x_{0} + iy_{0})}{x - x_{0}}$
                $= \displaystyle \lim_{x \to x_{0}} \frac{u(x, y_{0}) - u(x_{0}, y_{0})}{x - x_{0}} + i \frac{v(x, y_{0}) - v(x_{0}, y_{0})}{x - x_{0}} $
                $= \displaystyle (\frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x})(x_{0}, y_{0})$
                $= \displaystyle \frac{\partial (u + iv)}{\partial x}(x_{0}, y_{0})$
                $= \displaystyle \frac{\partial f}{\partial x}(x_{0}, y_{0}) $

 

  Also

  $f'(z_{0}) = \displaystyle \lim_{z \to z_{0}} \frac{f(z) - f(z_{0})}{z - z_{0}}$
                $= \displaystyle \lim_{x_{0} + iy \to z_{0}} \frac{f(x_{0} + iy) - f(z_{0})}{x_{0} + iy - z_{0}}$
                $= \displaystyle \lim_{y \to y_{0}} \frac{f(x_{0} + iy) - f(x_{0} + iy_{0})}{i(y - y_{0})}$
                $= \displaystyle \lim_{y \to y_{0}} -i[\frac{u(x_{0}, y) - u(x_{0}, y_{0})}{y - y_{0}} + i \frac{v(x_{0}, y) - v(x_{0}, y_{0})}{y - y_{0}} ]$
                $= \displaystyle -i(\frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y})(x_{0}, y_{0})$
                $= \displaystyle -i\frac{\partial (u + iv)}{\partial y}(x_{0}, y_{0})$
                $= \displaystyle -i\frac{\partial f}{\partial y}(x_{0}, y_{0}) $

 

  $\therefore$  $\displaystyle \frac{\partial f}{\partial x}(x_{0}, y_{0}) = -i \frac{\partial f}{\partial y}(x_{0}, y_{0})$

 

 

 

Cor 1

$f(z) = u(x,y) + i v(x, y)$ is differentiable at $z_{0}$

$\Rightarrow$  $f'(z_{0}) = \dfrac{\partial f}{\partial x} (z_{0}) = - i \dfrac{\partial f}{\partial y} (z_{0})$

 

 

 

Thm 3

For $f(z) = u(x,y) + i v(x, y)$,  C-R equation is satisfied for all $z$

$\Rightarrow$  $\Delta u = 0$  and  $\Delta v = 0$

( $\Delta$ is Laplace operator )

 

  $\Delta u = \dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2}$
           $= \dfrac{\partial }{\partial x} (\dfrac{\partial u}{\partial x}) + \dfrac{\partial }{\partial y} (\dfrac{\partial u}{\partial y})$
           $= \dfrac{\partial }{\partial x} (\dfrac{\partial v}{\partial y}) + \dfrac{\partial }{\partial y} (-\dfrac{\partial v}{\partial x})$
           $= \dfrac{\partial^{2} v}{\partial x \partial y} - \dfrac{\partial^{2} v}{\partial y \partial x} = 0 $

 

  Similarly $\Delta v = 0$

 

 

 

Cor 2

$f(z) = u(x,y) + i v(x, y)$ is differentiable at all points.

$\Rightarrow$  $\Delta u = 0$  and  $\Delta v = 0$

( $\Delta$ is Laplace operator )