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
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