Difference of $\mathbb{C}$ and $\mathbb{R}^{2}$

 

Def 1

$u : \mathbb{R}^{2} \to \mathbb{R}$ differentaible at $(x_{0}, y_{0})$

$\Leftrightarrow$  $\exists \dfrac{\partial u}{\partial x}(x_{0}, y_{0}), \dfrac{\partial u}{\partial y}(x_{0}, y_{0})$  and  $\displaystyle \frac{\left|u(x_{0} + h, y_{0} + k) - [u(x_{0}, y_{0}) + \frac{\partial u}{\partial x}(x_{0}, y_{0})h + \frac{\partial u}{\partial y}(x_{0}, y_{0})k] \right|}{\left|(h, k) \right|} \to 0$  as  $\left|(h, k) \right| \to 0$

 

 

 

Thm 1

Let $u : \mathbb{R}^{2} \to \mathbb{R}$

The partial derivatives of $u$ exist in the neighborhood of $(x_{0}, y_{0})$ and are continuous at $(x_{0}, y_{0})$

$\Rightarrow$  $u$ is differentaible at $(x_{0}, y_{0})$

 

  $\forall \varepsilon > 0$

 

  Since partial derivative of $u$ exists in the neighborhood of $(x_{0}, y_{0})$, there exists $\delta_{1} > 0$  s.t.

  $\left|(x, y) - (x_{0}, y_{0}) \right| < \delta_{1} $  $\Rightarrow$  $\exists \dfrac{\partial u}{\partial x}(x, y)$,  $\dfrac{\partial u}{\partial y}(x, y)$

  Since $\dfrac{\partial u}{\partial x}$ is continuous at $(x_{0}, y_{0})$, there exists $\delta_{2} >0$  s.t.

  $\left|(x, y) - (x_{0}, y_{0}) \right| < \delta_{2} $  $\Rightarrow$  $\left|\dfrac{\partial u}{\partial x}(x, y) - \dfrac{\partial u}{\partial x}(x_{0}, y_{0}) \right| < \dfrac{\varepsilon}{\sqrt{2}}$

  Since $\dfrac{\partial u}{\partial y}$ is continuous at $(x_{0}, y_{0})$, there exists $\delta_{3} >0$  s.t.

  $\left|(x, y) - (x_{0}, y_{0}) \right| < \delta_{3} $  $\Rightarrow$  $\left|\dfrac{\partial u}{\partial y}(x, y) - \dfrac{\partial u}{\partial y}(x_{0}, y_{0}) \right| < \dfrac{\varepsilon}{\sqrt{2}}$

 

  Let $\delta = \min \left\{\delta_{1}, \delta_{2}, \delta_{3} \right\} $

 

 

  Suppose : $0 < \left|(h, k) \right| < \delta$

  By mean value theorem,

  $u(x_{0} + h, y_{0} + k) - u(x_{0}, y_{0})$

       $= u(x_{0} + h, y_{0} + k) - u(x_{0}, y_{0} + k) + u(x_{0}, y_{0} + k) - u(x_{0}, y_{0})$
       $= h \dfrac{\partial u}{\partial x} (x_{0} + h', y_{0} + k) + k \dfrac{\partial u}{\partial y} (x_{0}, y_{0} + k')$

  ( $h'$ lies between $x_{0}$, $x_{0} + h$  and  $k'$ lies between $y_{0}$, $y_{0} + k$ )

 

  By Cauchy-Schwarz inequality,

  $\left|u(x_{0} + h, y_{0} + k) - u(x_{0}, y_{0}) - h \dfrac{\partial u}{\partial x} (x_{0}, y_{0}) - k \dfrac{\partial u}{\partial y} (x_{0}, y_{0}) \right|$

    $= \left|h (\dfrac{\partial u}{\partial x} (x_{0} + h', y_{0} + k) - \dfrac{\partial u}{\partial x} (x_{0}, y_{0})) + k (\dfrac{\partial u}{\partial y} (x_{0}, y_{0} + k') - \dfrac{\partial u}{\partial y} (x_{0}, y_{0})) \right|$

    $\leq \sqrt{h^{2} + k^{2}} \sqrt{(\dfrac{\partial u}{\partial x} (x_{0} + h', y_{0} + k) - \dfrac{\partial u}{\partial x} (x_{0}, y_{0}))^{2} + (\dfrac{\partial u}{\partial y} (x_{0}, y_{0} + k') - \dfrac{\partial u}{\partial y} (x_{0}, y_{0}))^{2}}$

 

  Therefore

  $\displaystyle \frac{\left|u(x_{0} + h, y_{0} + k) - [u(x_{0}, y_{0}) + \frac{\partial u}{\partial x}(x_{0}, y_{0})h + \frac{\partial u}{\partial y}(x_{0}, y_{0})k] \right|}{\left|(h, k) \right|}$
       $\leq \sqrt{(\dfrac{\partial u}{\partial x} (x_{0} + h', y_{0} + k) - \dfrac{\partial u}{\partial x} (x_{0}, y_{0}))^{2} + (\dfrac{\partial u}{\partial y} (x_{0}, y_{0} + k') - \dfrac{\partial u}{\partial y} (x_{0}, y_{0}))^{2}}$

       $< \sqrt{(\dfrac{\varepsilon}{\sqrt{2}})^{2} + (\dfrac{\varepsilon}{\sqrt{2}})^{2}} = \varepsilon$

 

  end

 

  $\therefore$  $\displaystyle \lim_{\left|(h, k) \right| \to 0}\frac{\left|u(x_{0} + h, y_{0} + k) - [u(x_{0}, y_{0}) + \frac{\partial u}{\partial x}(x_{0}, y_{0})h + \frac{\partial u}{\partial y}(x_{0}, y_{0})k] \right|}{\left|(h, k) \right|} = 0$

  $\therefore$  $u$ is differentiable at $(x_{0}, y_{0})$

 

 

 

Thm 2

For $f(z) = u(x, y) + iv(x, y)$,

$u, v$ are differentiable at $z_{0} = x_{0} + iy_{0}$ .

The C-R equation of $f$ is satisfied at $z_{0}$ .

$\Rightarrow$  $f$ is differentiable at $z_{0}$

 

  Let

$$u(x_{0} + h, y_{0} + k) - u(x_{0}, y_{0}) - h \frac{\partial u}{\partial x} (x_{0}, y_{0}) - k \frac{\partial u}{\partial y} (x_{0}, y_{0}) = e_{1}(h, k)$$

$$v(x_{0} + h, y_{0} + k) - v(x_{0}, y_{0}) - h \frac{\partial v}{\partial x} (x_{0}, y_{0}) - k \frac{\partial v}{\partial y} (x_{0}, y_{0}) = e_{2}(h, k)$$

 

  Since $u, v$ are differentiable at $(x_{0}, y_{0})$,

  $\displaystyle \lim_{\left|(h,k) \right| \to 0} \frac{\left| e_{i}(h, k) \right|}{\left|(h, k) \right|} = 0$     for  $i = 1, 2$

 

  Let $l = h+ik$

 

  $\dfrac{f(z_{0} + l) - f(z_{0})}{l}$
      $= \displaystyle \frac{1}{l} ([h \frac{\partial u}{\partial x}(x_{0}, y_{0}) + k \frac{\partial u}{\partial y}(x_{0}, y_{0}) + e_{1}] + i[h \frac{\partial v}{\partial x}(x_{0}, y_{0}) + k \frac{\partial v}{\partial y}(x_{0}, y_{0}) + e_{2}])$
      $= \displaystyle \frac{1}{l}(e_{1} + i e_{2}) + \frac{1}{l} (h \frac{\partial u}{\partial x}(x_{0}, y_{0}) + i k \frac{\partial v}{\partial y}(x_{0}, y_{0})) + \frac{i}{l} (h \frac{\partial v}{\partial x}(x_{0}, y_{0}) - i k \frac{\partial u}{\partial y}(x_{0}, y_{0}))$
      $= \displaystyle \frac{1}{l}(e_{1} + i e_{2}) + \frac{1}{l} (h \frac{\partial u}{\partial x}(x_{0}, y_{0}) + i k \frac{\partial u}{\partial x}(x_{0}, y_{0})) + \frac{i}{l} (h \frac{\partial v}{\partial x}(x_{0}, y_{0}) + i k \frac{\partial v}{\partial x}(x_{0}, y_{0}))$
       $= \displaystyle \frac{1}{l}(e_{1} + i e_{2}) + \frac{\partial u}{\partial x}(x_{0}, y_{0}) + i \frac{\partial v}{\partial x}(x_{0}, y_{0})$

      $= \displaystyle \frac{1}{l}(e_{1} + i e_{2}) + \frac{\partial f}{\partial x}(x_{0}, y_{0})$

 

Since $\displaystyle \lim_{\left|l \right| \to 0} \left| \frac{e_{i}}{l} \right| = 0$  for  $i = 1, 2$,

$\displaystyle \lim_{l \to 0} \frac{e_{i}}{l} = 0$     for  $i=1, 2$

$\therefore$  $\displaystyle \lim_{l \to 0} \frac{e_{1} + ie_{2}}{l} = 0$

$\therefore$  $\displaystyle \lim_{l \to 0} \frac{f(z_{0} + l) - f(z_{0})}{l} = \frac{\partial f}{\partial x}(z_{0})$

$\therefore$  $f(z)$ is differentiable at $z_{0}$

 

 

 

Cor 1

For $f(z) = u(x, y) + iv(x, y)$,

The partial derivatives of $u, v$ exist in the neighborhood of $z_{0} = x_{0} + iy_{0}$ and are continuous at $z_{0}$ .

The C-R equation of $f$ is satisfied at $z_{0}$ .

$\Rightarrow$  $f$ is differentiable at $z_{0}$

 

 

 

Thm 3

For $f(z) = u(x,y) + iv(x, y)$,

$f$ is differentiable at $z_{0} = x_{0} + iy_{0}$

$\Rightarrow$  $u$, $v$ are differentiable at $z_{0}$

 

  Show 1 : The partial derivatives of $u$, $v$ exist at $z_{0}$

  By Thm 1, it is clear!

 

  end

 

 

  Show 2 : $\displaystyle \lim_{\left|(h, k) \right| \to 0} \dfrac{\left| u(x_{0} + h, y_{0} + k) - u(x_{0}, y_{0}) - \dfrac{\partial u}{\partial x}(x_{0}, y_{0})h - \dfrac{\partial u}{\partial y}(x_{0}, y_{0})k \right|}{\left|(h, k) \right|} =0$

  $\forall \varepsilon > 0$

 

  Since $f$ is differentiable at $z_{0}$, there exists $\delta >0$  s.t.

  $0 < \left|l \right| < \delta$  $\Rightarrow$  $\left|\dfrac{f(z_{0} + l) - f(z_{0})}{l} - f'(z_{0}) \right| < \varepsilon$

 

  Let $l = h + ik$ and $f'(z_{0}) = a + ib$

 

  $\left|\dfrac{f(z_{0} + l) - f(z_{0})}{l} - f'(z_{0}) \right|$

  $= \left|\dfrac{u(x_{0} + h, y_{0} + k) + iv(x_{0} + h, y_{0} + k) - u(x_{0}, y_{0}) - iv(x_{0}, y_{0})}{h + ik} - a - ib \right| $

  $= \left|\dfrac{u(x_{0} + h, y_{0} + k) + iv(x_{0} + h, y_{0} + k) - u(x_{0}, y_{0}) - iv(x_{0}, y_{0}) - ah + bk - ibh - iak}{h + ik} \right| $

  $= \dfrac{\left| [u(x_{0} + h, y_{0} + k) - u(x_{0}, y_{0}) - ah + bk] + i[v(x_{0} + h, y_{0} + k) - v(x_{0}, y_{0}) - bh - ak] \right|}{\left|(h, k) \right|} $

 

  Therefore

  $0 < \left|l \right| < \delta$

  $\Rightarrow$  $\dfrac{\left| u(x_{0} + h, y_{0} + k) - u(x_{0}, y_{0}) - ah + bk \right|}{\left|(h, k) \right|}$

  $\leq \dfrac{\left| [u(x_{0} + h, y_{0} + k) - u(x_{0}, y_{0}) - ah + bk] + i[v(x_{0} + h, y_{0} + k) - v(x_{0}, y_{0}) - bh - ak] \right|}{\left|(h, k) \right|}$

  $< \varepsilon$

 

  $\therefore$  $\displaystyle \lim_{\left|(h, k) \right| \to 0} \dfrac{\left| u(x_{0} + h, y_{0} + k) - u(x_{0}, y_{0}) - ah + bk \right|}{\left|(h, k) \right|} =0$

 

 

  By Cor 1,

  $f'(z_{0}) = \dfrac{\partial f}{\partial x}(z_{0}) = \dfrac{\partial u}{\partial x}(x_{0}, y_{0}) + i\dfrac{\partial v}{\partial x}(x_{0}, y_{0})$

 

  $\therefore$  $a = \dfrac{\partial u}{\partial x}(x_{0}, y_{0})$,  $b = \dfrac{\partial v}{\partial x}(x_{0}, y_{0})$

 

  By Thm 2,

  C-R equation of $f$ is satisfied at $z_{0}$

  $\therefore$  $b = -\dfrac{\partial u}{\partial y}(x_{0}, y_{0})$

 

  $\therefore$  $\displaystyle \lim_{\left|(h, k) \right| \to 0} \dfrac{\left| u(x_{0} + h, y_{0} + k) - u(x_{0}, y_{0}) - \dfrac{\partial u}{\partial x}(x_{0}, y_{0})h - \dfrac{\partial u}{\partial y}(x_{0}, y_{0})k \right|}{\left|(h, k) \right|} =0$

 

  end

 

 

  Show 3 : $\displaystyle \lim_{\left|(h, k) \right| \to 0} \dfrac{\left| v(x_{0} + h, y_{0} + k) - v(x_{0}, y_{0}) - \dfrac{\partial v}{\partial x}(x_{0}, y_{0})h - \dfrac{\partial v}{\partial y}(x_{0}, y_{0})k \right|}{\left|(h, k) \right|} =0$

  This is similar to Show 2.

 

  end

 

  $\therefore$  $u$, $v$ are diffrenetiable at $z_{0}$

 

 

 

Cor 2

For $f(z) = u(x,y) + iv(x, y)$,

$f$ is differentiable at $z_{0} = x_{0} + iy_{0}$

$\Leftrightarrow$  $u$, $v$ are differentiable at $z_{0}$     &     C-R equation of $f$ is satisfied at $z_{0}$