Line integral of complex function

 

Def 1

Let $\Omega$ be an area and $\Gamma : [a, b] \to \Omega$ . 

1.  If both the real and imaginary parts of $\Gamma$ are $C^{1}$-function, it is called a $C^{1}$-curve.

2.  A curve that divides the interval $[a, b]$ into a finite number of subintervals, where it is a $C^{1}$-curve on each subinterval and continuous over the entire interval $[a, b]$, is called a piecewise $C^{1}$-curve.

 

# From now on, the term 'curve' will always refer to a piecewise $C^{1}$-curve.

 

 

 

Def 2

Let $\Omega$ be an area,  $\Gamma : [a, b] \to \Omega$ be a curve  and  $f : \Omega \to \mathbb{C}$

$\displaystyle \int_{\Gamma} f(z) \; \mathrm{d}z = \int_{a}^{b} f(\Gamma(t)) \Gamma'(t) \; \mathrm{d}t$

 

 

 

Prop 1

Let $\Omega$ be an area,  $\Gamma : [a, b] \to \Omega$ be a curve  and  $f : \Omega \to \mathbb{C}$

Let $\Gamma(t) = x(t) + i y(t)$  and  $f(z) = u(x,y) + i v(x,y)$ . Then

$\displaystyle \int_{\Gamma} f(z) \; \mathrm{d}z = (\int_{\Gamma} u \; \mathrm{d}x - v \; \mathrm{d}y) + i (\int_{\Gamma} v \; \mathrm{d}x + u \; \mathrm{d}y)$

 

  $\displaystyle \int_{\Gamma} f(z) \; \mathrm{d}z = \int_{a}^{b} f(\Gamma(t)) \Gamma'(t) \; \mathrm{d}t$

                        $\displaystyle = \int_{a}^{b} (u+iv)(x' + iy') \; \mathrm{d}t$

                        $\displaystyle = \int_{a}^{b} (ux' - vy') \; \mathrm{d}t + i \int_{a}^{b} (vx' + uy') \; \mathrm{d}t$

                        $\displaystyle = \int_{a}^{b} (u \dfrac{\mathrm{d} x}{\mathrm{d} t} - v \dfrac{\mathrm{d} y}{\mathrm{d} t}) \; \mathrm{d}t + i \int_{a}^{b} (v \dfrac{\mathrm{d} x}{\mathrm{d} t} + u \dfrac{\mathrm{d} y}{\mathrm{d} t}) \; \mathrm{d}t$

                        $\displaystyle = (\int_{\Gamma} u \; \mathrm{d}x - v \; \mathrm{d}y) + i (\int_{\Gamma} v \; \mathrm{d}x + u \; \mathrm{d}y)$

 

 

 

Def 3

Let $\Gamma_{i} = [a_{i}, \; b_{i}] \to \mathbb{C}$ be a curve for $i = 1, 2, \cdots, n$ .

Let $\Gamma_{i} (b_{i}) = \Gamma_{i+1} (a_{i+1})$  for $i = 1, 2, \cdots, n-1$ .

 

Define $\displaystyle \sum_{i=1}^{n} \Gamma_{i} : [a_{1}, \; b_{1} + \sum_{i=2}^{n} (b_{i} - a_{i})] \to \mathbb{C}$ by

$(\displaystyle \sum_{i=1}^{n} \Gamma_{i}) (t) = \begin{cases}
\Gamma_{1} (t) & t \in [a_{1}, \; b_{1}] \\
\Gamma_{2} (t + (a_{2} - b_{1})) & t \in [b_{1}, \; b_{1} + (b_{2} - a_{2})] \\
\vdots &  \\
\Gamma_{n} (t + (a_{n} - b_{1}) - \displaystyle \sum_{i=2}^{n-1}(b_{i} - a_{i})) & t \in [b_{1} + \displaystyle \sum_{i=2}^{n-1}(b_{i} - a_{i}), \; b_{1} + \displaystyle \sum_{i=2}^{n}(b_{i} - a_{i})] 
\end{cases}$

 

 

 

Prop 2

Let $\Omega$ be an area  and  $\Gamma_{i} = [a_{i}, \; b_{i}] \to \Omega$ be a curve for $i = 1, 2, \cdots, n$ .

Let $\Gamma_{i} (b_{i}) = \Gamma_{i+1} (a_{i+1})$  for $i = 1, 2, \cdots, n-1$ .

Let $f: \Omega \to \mathbb{C}$ .

  $ \displaystyle \int_{ \sum_{i=1}^{n} \Gamma_{i}} f(z) \; \mathrm{d}z = \sum_{i=1}^{n} \int_{\Gamma_{i}} f(z) \; \mathrm{d}z$

 

  We will use an induction argument to prove

  $ \displaystyle \int_{ \sum_{i=1}^{n} \Gamma_{i}} f(z) \; \mathrm{d}z = \sum_{i=1}^{n} \int_{\Gamma_{i}} f(z) \; \mathrm{d}z$

 

  Show 1 : holds for $n = 1$

  It is clear.

 

  end

 

 

  Show 2 : If holds for $n = k$, holds for $n = k+1$

  Suppose  $ \displaystyle \int_{ \sum_{i=1}^{k} \Gamma_{i}} f(z) \; \mathrm{d}z = \sum_{i=1}^{k} \int_{\Gamma_{i}} f(z) \; \mathrm{d}z$

 

  $\displaystyle \int_{ \sum_{i=1}^{k+1} \Gamma_{i}} f(z) \; \mathrm{d}z$

       $= \displaystyle \int_{a_{1}}^{b_{1} + \sum_{i=2}^{k+1} (b_{i} - a_{i})} f( (\sum_{i=1}^{k+1} \Gamma_{i})(t) ) (\sum_{i=1}^{k+1} \Gamma_{i})'(t) \; \mathrm{d}t $
       $= \displaystyle \int_{a_{1}}^{b_{1} + \sum_{i=2}^{k} (b_{i} - a_{i})} f( (\sum_{i=1}^{k+1} \Gamma_{i})(t) ) (\sum_{i=1}^{k+1} \Gamma_{i})'(t) \; \mathrm{d}t$

                 $\displaystyle + \int_{b_{1} + \sum_{i=2}^{k} (b_{i} - a_{i})}^{b_{1} + \sum_{i=2}^{k+1} (b_{i} - a_{i})} f( (\sum_{i=1}^{k+1} \Gamma_{i})(t) ) (\sum_{i=1}^{k+1} \Gamma_{i})'(t) \; \mathrm{d}t$

       $= \displaystyle \int_{ \sum_{i=1}^{k} \Gamma_{i}} f(z) \; \mathrm{d}z + \int_{b_{1} + \sum_{i=2}^{k} (b_{i} - a_{i})}^{b_{1} + \sum_{i=2}^{k+1} (b_{i} - a_{i})} f( \Gamma_{k+1}(t + (a_{k+1} - b_{1}) - \displaystyle \sum_{i=2}^{k}(b_{i} - a_{i})))$

                 $\displaystyle \Gamma_{k+1}'(t + (a_{k+1} - b_{1}) - \displaystyle \sum_{i=2}^{k}(b_{i} - a_{i})) \; \mathrm{d}t $

       $= \displaystyle \sum_{i=1}^{k} \int_{\Gamma_{i}} f(z) \; \mathrm{d}z + \int_{a_{k+1}}^{b_{k+1}} f( \Gamma_{k+1}(s) ) \Gamma_{k+1}'(s) \; \mathrm{d}s $

       $= \displaystyle \sum_{i=1}^{k} \int_{\Gamma_{i}} f(z) \; \mathrm{d}z + \int_{\Gamma_{k+1}} f(z) \; \mathrm{d}z$
       $= \displaystyle \sum_{i=1}^{k+1} \int_{\Gamma_{i}} f(z) \; \mathrm{d}z $

 

  end

 

 

 

Def 4

Let $\Omega$ be an area  and  $\Gamma = [a, b] \to \mathbb{C}$ be a curve.

Define $- \Gamma : [a, b] \to \mathbb{C}$ by

$(- \Gamma) (t) = \Gamma(b + a - t)$     $t \in [a, b]$

 

 

 

Prop 3

Let $\Omega$ be an area,  $\Gamma = [a, b] \to \Omega$ be a curve  and  $f : \Omega \to \mathbb{C}$

$\displaystyle \int_{- \Gamma} f(z) \; \mathrm{d}z = - \int_{\Gamma} f(z) \; \mathrm{d}z$

 

  $\displaystyle \int_{- \Gamma} f(z) \; \mathrm{d}z = \int_{a}^{b} f((-\Gamma)(t)) \; (-\Gamma)'(t) \; \mathrm{d}t$
                           $= - \displaystyle \int_{a}^{b} f(\Gamma(b + a - t)) \; \Gamma'(b + a - t) \; \mathrm{d}t$
                           $= \displaystyle \int_{b}^{a} f(\Gamma(s)) \; \Gamma'(s) \; \mathrm{d}s$
                           $= - \displaystyle \int_{\Gamma} f(z) \; \mathrm{d}z$

 

 

 

Prop 4

Let $\Omega$ be an area,  $\Gamma : [a, b] \to \Omega$ be a curve  and  $f : \Omega \to \mathbb{C}$

Let $g : [c, d] \to [a,b]$ be 1-1 onto increasing function  and  $\gamma = \Gamma \circ g$

$\displaystyle \int_{\gamma} f \; \mathrm{d}z = \int_{\Gamma} f \; \mathrm{d}z$

 

  $\displaystyle \int_{\gamma} f \; \mathrm{d}z = \int_{c}^{d} f(\gamma(t)) \gamma'(t) \; \mathrm{d}t$
                  $= \displaystyle \int_{c}^{d} f(\Gamma (g(t)) ) \Gamma'(g(t)) g'(t) \; \mathrm{d}t$
                  $= \displaystyle \int_{a}^{b} f(\Gamma (s) ) \Gamma'(s) \; \mathrm{d}s$
                  $= \displaystyle \int_{\Gamma} f \; \mathrm{d}z $