복소함수 미분

 

Def 1

Let $\Omega \subset \mathbb{C}$ . $\Omega$ is an area of compex plane if

1.  $\forall z \in \Omega$, there exists $r > 0$  s.t.  $D(z, r) \subset \Omega$

2.  Any two points $z, w \in \Omega$ can be connected by a finite number of line segments within $\Omega$, each of which is either parallel to the real axis or the imaginary axis.

 

 

 

Def 2 

Let $\Omega$ be an area and $f : \Omega \to \mathbb{C}$

For $z_{0} \in \Omega$, we define

$\displaystyle \lim_{z \to z_{0}} f(z) = \alpha $ 

$\Leftrightarrow$  $\displaystyle \lim_{\left|z - z_{0} \right| \to 0} \left|f(z) - \alpha \right| = 0 $

$\Leftrightarrow$  $\forall \varepsilon > 0, \; \exists \delta >0$  s.t.  $0 < \left|z - z_{0} \right| < \delta \Rightarrow \left|f(z) - \alpha \right| < \varepsilon $

 

 

 

Prop 1

$\displaystyle \lim_{z \to z_{0}} f(z) = \alpha$ 

$\Leftrightarrow$  $\displaystyle \lim_{z \to z_{0}} \mathrm{Re} \; f(z) = \mathrm{Re} \; \alpha $,     $\displaystyle \lim_{z \to z_{0}} \mathrm{Im} \; f(z) = \mathrm{Im} \; \alpha$

 

pf)

$\Rightarrow)$

  $\forall \varepsilon > 0$, there exists $\delta>0$  s.t.

  $0 < \left|z - z_{0} \right| < \delta$  $\Rightarrow$  $\left|f(z) - \alpha \right| < \varepsilon $

  Since $\left| \mathrm{Re} \; f(z) - \mathrm{Re} \; \alpha \right| \leq \left|f(z) - \alpha \right|$,

  $0 < \left|z - z_{0} \right| < \delta$  $\Rightarrow$  $\left|\mathrm{Re} \; f(z) - \mathrm{Re} \; \alpha \right| < \varepsilon $

  $\therefore$  $\displaystyle \lim_{z \to z_{0}} \mathrm{Re} \; f(z) = \mathrm{Re} \; \alpha$

  Similarly, We can also show $\displaystyle \lim_{z \to z_{0}} \mathrm{Im} \; f(z) = \mathrm{Im} \; \alpha$

 

$\Leftarrow)$

  $\forall \varepsilon > 0$, there exists $\delta_{1}, \delta_{2} >0$  s.t.

  $0 < \left|z - z_{0} \right| < \delta_{1}$  $\Rightarrow$  $\left|\mathrm{Re} \; f(z) - \mathrm{Re} \; \alpha \right| < \dfrac{\varepsilon}{2} $

  $0 < \left|z - z_{0} \right| < \delta_{2}$  $\Rightarrow$  $\left|\mathrm{Im} \; f(z) - \mathrm{Im} \; \alpha \right| < \dfrac{\varepsilon}{2} $

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

  $0 < \left|z - z_{0} \right| < \delta$  $\Rightarrow$  $\left|f(z) - \alpha \right| = \left|\mathrm{Re} \; f(z) - \mathrm{Re} \; \alpha + i(\mathrm{Im} \; f(z) - \mathrm{Im} \; \alpha) \right|$

                                                                    $\leq \left|\mathrm{Re} \; f(z) - \mathrm{Re} \; \alpha \right| + \left| \mathrm{Im} \; f(z) - \mathrm{Im} \; \alpha \right|$

                                                                    $< \dfrac{\varepsilon}{2} + \dfrac{\varepsilon}{2} = \varepsilon$

  $\therefore$  $\displaystyle \lim_{z \to z_{0}} f(z) = \alpha$

 

 

 

Def 3

1.  $f(z)$ is differentiable at $z = z_{0}$  $\Leftrightarrow$  $\exists \displaystyle \lim_{z \to z_{0}} \frac{f(z) - f(z_{0})}{z - z_{0}}$

2.  $f'(z_{0}) = \displaystyle \lim_{z \to z_{0}} \frac{f(z) - f(z_{0})}{z - z_{0}}$

3.  $f(z)$ is differentiable at all points  $\Rightarrow$  $f'(z)$ called derivate of $f(z)$

 

 

 

Prop2

Let $f, g$ are differentiable.

1.  $(f + g)' = f' + g'$

2.  $(fg)' = f'g + fg'$

 

pf)

  $\forall z_{0} \in \mathbb{C}$

  $(f + g)'(z_{0}) = \displaystyle \lim_{z \to z_{0}} \frac{(f(z) + g(z)) - (f(z_{0}) + g(z_{0}))}{z - z_{0}}$

                            $= \displaystyle \lim_{z \to z_{0}} \frac{f(z) - f(z_{0})}{z - z_{0}} + \displaystyle \lim_{z \to z_{0}} \frac{g(z) - g(z_{0})}{z - z_{0}}$

                            $ = f'(z_{0}) + g'(z_{0})$

                            $ = (f' + g')(z_{0})$

$\therefore$  $(f+g)' = f' + g'$

 

  $\forall z_{0} \in \mathbb{C}$

  $(fg)'(z_{0}) = \displaystyle \lim_{z \to z_{0}} \frac{f(z)g(z) - f(z_{0})g(z_{0})}{z - z_{0}} $
                      $= \displaystyle \lim_{z \to z_{0}} \frac{f(z)g(z) - f(z_{0})g(z) + f(z_{0})g(z) - f(z_{0})g(z_{0})}{z - z_{0}} $
                      $= \displaystyle \lim_{z \to z_{0}} \frac{f(z) - f(z_{0})}{z - z_{0}} \cdot g(z) + \displaystyle \lim_{z \to z_{0}} f(z_{0}) \cdot \frac{f(z) - f(z_{0})}{z - z_{0}} $
                      $= f'(z_{0})g(z_{0}) + f(z_{0})g'(z_{0})$
                      $= (f'g + fg')(z_{0})$

  $\therefore$  $(fg)' = f'g + fg'$

 

 

 

Prop 3

$f(z) = z^{n}$  $\Rightarrow$  $f'(z) = n z^{n-1}$

 

  $\forall z_{0} \in \mathbb{C}$

$$f'(z_{0}) = \displaystyle \lim_{z \to z_{0}} \dfrac{z^{n} - z_{0}^{n}}{z - z_{0}} = \displaystyle \lim_{z \to z_{0}} (z^{n-1} + z^{n-2}z_{0} + \cdots + z_{0}^{n-1})= n z_{0}^{n-1}$$

  $\therefore$  $f'(z) = nz^{n-1}$

 

 

 

Thm 1

Let $\Omega$ be an area and $f : \Omega \to \mathbb{C}$

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

$\Leftrightarrow$  there eixsts $\alpha \in \mathbb{C}$,  function $h : \Omega \setminus \left\{z_{0} \right\} \to \mathbb{C}$  s.t.

$f(z) = f(z_{0}) + \alpha (z - z_{0}) + h(z) (z - z_{0})$     &     $\displaystyle \lim_{z \to z_{0}} h(z) = 0$

 

# In this case $\alpha = f'(z_{0})$

 

pf)

$\Rightarrow)$

  Let $h(z) = \dfrac{f(z) - f(z_{0})}{z - z_{0}} - f'(z_{0})$

  Since $f(z)$ is differentiable at $z_{0}$,

  $\displaystyle \lim_{z \to z_{0}} h(z) = 0$

 

  Let $\alpha = f'(z_{0})$

  $\therefore$  $f(z) = f(z_{0}) + \alpha (z - z_{0}) + h(z) (z - z_{0})$

 

$\Leftarrow)$

  Since $\displaystyle \lim_{z \to z_{0}} h(z) = 0$,

  $\displaystyle \lim_{z \to z_{0}} \dfrac{f(z) - f(z_{0})}{z - z_{0}} = \displaystyle \lim_{z \to z_{0}} \alpha + h(z) = \alpha$

 

  $\therefore$  $f(z)$ is differentiable at $z_{0}$  and  $\alpha = f'(z_{0})$

 

 

 

Thm 2

Let $\Omega$ be an area and $f : \Omega \to \mathbb{C}$

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

$\Rightarrow$  $f(z)$ is continuous at $z_{0}$

 

  Since $f$ is differentiable at $z_{0}$,

  $\displaystyle \lim_{z \to z_{0}} f(z) - f(z_{0}) = \displaystyle \lim_{z \to z_{0}} \dfrac{f(z) - f(z_{0})}{z-z_{0}} \cdot (z - z_{0}) = f'(z_{0}) \cdot 0 = 0$

  $\therefore$  $\displaystyle \lim_{z \to z_{0}} f(z) = f(z_{0})$

 

 

 

Prop 4

Let $f, g$ are differentiable at $z_{0}$ . Then

$(\dfrac{f}{g})'(z_{0}) = (\dfrac{f'g - fg'}{g^{2}}) (z_{0})$     ( $g(z_{0}) \neq 0$ )

 

  $\dfrac{(f/g)(x) - (f/g)(c)}{x - c} = \dfrac{f(x)g(c) - f(c)g(x)}{(x - c)g(x)g(c)}$

                                                $= \dfrac{f(x)g(c) - f(c)g(c) + f(c)g(c) - f(c)g(x)}{(x - c)g(x)g(c)}$

                                                $= \dfrac{1}{g(x)g(c)}(\dfrac{f(x) - f(c)}{x - c}g(c) - \dfrac{g(x) - g(c)}{x - c}f(c))$

 

  Since $g$ is differentiable at $z_{0}$ and Thm 2,

  $g$ is continuous at $z_{0}$

 

  Since $g(z_{0}) \neq 0$, there exists $\delta > 0$  s.t.

  $0 < \left|z - z_{0} \right| < \delta$  $\Rightarrow$  $0 < \left|g(z) \right|$

 

  $\therefore$  $(\dfrac{f}{g})'(z_{0}) = (\dfrac{f'g - fg'}{g^{2}}) (z_{0})$

 

 

 

Prop 5

Let $\Omega_{1}, \Omega_{2}$ be an area,  $f : \Omega_{1} \to \mathbb{C}$,  $g : \Omega_{2} \to \mathbb{C}$  and  $f(\Omega_{1}) \subset \Omega_{2}$ . 

Let $f$ is differentiable at $z_{0}$  and  $g$ is differentiable at $f(z_{0})$ . Then

$(g \circ f)'(z_{0}) = g' (f(z_{0})) f'(z_{0})$

 

  Since $f$ is differentiable at $z_{0}$ and Thm 1, there exists $h_{f} : \Omega_{1} \setminus \left\{z_{0} \right\} \to \mathbb{C}$  s.t.

  $f(z) - f(z_{0}) = f'(z_{0}) (z - z_{0}) + h_{f}(z) (z - z_{0})$     &     $\displaystyle \lim_{z \to z_{0}} h_{f}(z) = 0$

 

  Also since $g$ is differentiable at $f(z_{0})$, there exists $h_{g} : \Omega_{2} \setminus \left\{f(z_{0}) \right\} \to \mathbb{C}$  s.t.

  $g(z) - g(f(z_{0})) = g'(f(z_{0})) (z - f(z_{0})) + h_{g}(z) (z - f(z_{0}))$     &     $\displaystyle \lim_{z \to f(z_{0})} h_{g}(z) = 0$

 

  Define $\overline{h}_{g} : \Omega_{2} \to \mathbb{C}$ by

  $\overline{h}_{g}(z) = \begin{cases}
h_{g}(z) & z \in \Omega_{2} \setminus \left\{f(z_{0}) \right\} \\
0 & z = f(z_{0}) 
\end{cases}$

  Since $\displaystyle \lim_{z \to f(z_{0})} h_{g}(z) = 0$,

  $\overline{h}_{g}$ is continuous at $f(z_{0})$

 

  Therefore

  $g(f(z)) - g(f(z_{0})) = (f(z) - f(z_{0}))(g(f'(z_{0})) + \overline{h}_{g}(f(z)))$ 

                                          $= (z - z_{0})(f'(z_{0}) + h_{f}(z))(g(f'(z_{0})) + \overline{h}_{g}(f(z)))$ 

 

  $\therefore$   $\dfrac{g(f(z)) - g(f(z_{0}))}{z - z_{0}} = (f'(z_{0}) + h_{f}(z))(g(f'(z_{0})) + \overline{h}_{g}(f(z))$

 

  Since $f$ is continuous at $z_{0}$ and $\overline{h}_{g}$ is continuous at $f(z_{0})$,

  $\displaystyle \lim_{z \to z_{0}} \overline{h}_{g}(f(z)) = \displaystyle \lim_{z \to f(z_{0})} \overline{h}_{g}(z) = 0$

 

  $\therefore$  $\displaystyle \lim_{z \to z_{0}} \dfrac{g(f(z)) - g(f(z_{0}))}{z - z_{0}} = g' (f(z_{0})) f'(z_{0})$

 

 

 

Def 4

Let $\Omega$ be an area and $f : \Omega \to \mathbb{C}$

For $z_{0} \in \Omega$, we define

$f(z)$ is analytic at $z = z_{0}$

$\Leftrightarrow$  $\exists r > 0$  s.t.  $a \in D(z_{0}, r)$  $\Rightarrow$  $f(z)$ is differentiable at $z = a$

$f(z)$ called analytic function if $f(z)$ is analytic at all points in $\Omega$

 

 

 

Cor 1

Let $\Omega$ be an area and $f : \Omega \to \mathbb{C}$

$f$ is differentiable at all points  $\Leftrightarrow$  $f$ is analytic function