Cauchy-Goursat Theorem
Thm 1 Cauchy-Goursat Theorem
Let $\Omega$ be an area, $R \subset \Omega$ be a rectangle and $f : \Omega \to \mathbb{C}$ be analytic function. Then
$\displaystyle \int_{\partial R} f(z) \; \mathrm{d}z = 0$
Divide a rectangle $R$ into four equal parts by bisecting each side of $R$,
and label these parts as $R^{1}$, $R^{2}$, $R^{3}$, and $R^{4}$.
By the cancellation of paths,
$\displaystyle \int_{\partial R} f(z) \; \mathrm{d}z = \sum_{i=1}^{4} \int_{\partial R^{i}} f(z) \; \mathrm{d}z$
$\therefore$ $\left|\displaystyle \int_{\partial R} f(z) \; \mathrm{d}z \right| \leq \displaystyle \sum_{i=1}^{4} \left|\int_{\partial R^{i}} f(z) \; \mathrm{d}z \right|$
At least one of the four rectangles ( let's call it $R_{1}$ ) must satisfy the following inequality.
$\left|\displaystyle \int_{\partial R_{1}} f(z) \; \mathrm{d}z \right| \geq \dfrac{1}{4} \left|\displaystyle \int_{\partial R} f(z) \; \mathrm{d}z \right|$
By repeating the same process on $R_{1}$, we obtain $R_{2}$ .
If we continue this process, we obtain a rectangle $R_{1} \supset R_{2} \supset \cdots \supset R_{n} \supset \cdots$ .
At this point, the following property is satisfied.
1. The rectangle $R_{n+1}$ is one of the four parts obtained by dividing rectangle $R_{n}$
into four equal parts.
2. $\left|\displaystyle \int_{\partial R_{n+1}} f(z) \; \mathrm{d}z \right| \geq \dfrac{1}{4} \left|\displaystyle \int_{\partial R_{n}} f(z) \; \mathrm{d}z \right|$
By nested intervals theorem,
it converges to a point $z_{0}$ in the rectangle $R$
Since $f$ is differentaible at $z = z_{0}$, there exists function $h(z)$ s.t.
$f(z) = f(z_{0}) + f'(z_{0})(z - z_{0}) + h(z)(z - z_{0})$ & $\displaystyle \lim_{z \to z_{0}} h(z) = 0$
By Cor 2,
$\dfrac{1}{4^{n}} \left|\displaystyle \int_{\partial R} f(z) \; \mathrm{d}z \right| \leq \left|\displaystyle \int_{\partial R_{n}} f(z) \; \mathrm{d}z \right| = \left| \displaystyle \int_{\partial R_{n}} (z - z_{0})h(z) \; \mathrm{d}z \right|$
Let the perimeter of the rectangle $R$ be denoted by $L$,
and the length of the diagonal of $R$ be denoted by $D$ .
So the perimeter of $R_{n}$ is $\dfrac{L}{2^{n}}$, and the length of the diagonal of $R_{n}$ is $\dfrac{D}{2^{n}}$ .
Therefore
$\left|z - z_{0} \right| \leq \dfrac{D}{2^{n}}$ for $z \in \partial R_{n}$
Since $h(z)$ is continuous on $\partial R_{n}$ and $\partial R_{n}$ is compact, there exists $M_{n} \geq 0$ s.t.
$M_{n} = \displaystyle \max_{z \in \partial R_{n}} \left|h(z) \right|$
By Thm 3,
$\dfrac{1}{4^{n}} \left|\displaystyle \int_{\partial R} f(z) \; \mathrm{d}z \right| \leq \left| \displaystyle \int_{\partial R_{n}} (z - z_{0})h(z) \; \mathrm{d}z \right| \leq \dfrac{D}{2^{n}} M_{n} \dfrac{L}{2^{n}}$
$\therefore$ $\left|\displaystyle \int_{\partial R} f(z) \; \mathrm{d}z \right| \leq DLM_{n}$
Since $\displaystyle \lim_{z \to z_{0}} h(z) = 0$,
$\displaystyle \lim_{n \to \infty} M_{n} = 0$
$\therefore$ $\displaystyle \int_{\partial R} f(z) \; \mathrm{d}z = 0$
Thm 2
Let $D = D(z_{0}, r)$ and $f : D \to \mathbb{C}$ be analytic function. Then
primitive function of $f$ exists
$\forall z \in D$
Let $z_{0} = x_{0} + iy_{0}$, $z = x + iy$
A rectangle is defined with $z_{0}$ and $z$ as its vertices.
Create a curve from $z_{0}$ to $z$ along the boundary of this rectangle.
Let the curve that first crosses a horizontal line be denoted as $\Gamma_{1}$,
and the curve that first crosses a vertical line be denoted as $\Gamma_{2}$ .
By Thm 1,
$\displaystyle \int_{\Gamma_{1} - \Gamma_{2}} f(z) \; \mathrm{d}z = 0$
$\therefore$ $\displaystyle \int_{\Gamma_{1}} f(z) \; \mathrm{d}z = \int_{\Gamma_{2}} f(z) \; \mathrm{d}z$
By Prop 2,
$\displaystyle \int_{\Gamma_{1}} f(z) \; \mathrm{d}z = \int_{x_{0}}^{x} f(t+iy_{0}) \; \mathrm{d}t + i\int_{y_{0}}^{y} f(x+it) \; \mathrm{d}t$
$\displaystyle \int_{\Gamma_{2}} f(z) \; \mathrm{d}z = i\int_{y_{0}}^{y} f(x_{0}+it) \; \mathrm{d}t + \int_{x_{0}}^{x} f(t+iy) \; \mathrm{d}t$
Let $\displaystyle F(z) = \int_{\Gamma_{1}} f(z) \; \mathrm{d}z = \int_{\Gamma_{2}} f(z) \; \mathrm{d}z$
$\displaystyle \frac{\partial F}{\partial y} = \frac{\partial }{\partial y} (i \int_{y_{0}}^{y} f(x+it) \; \mathrm{d}t) = if(x + iy) = if(z)$
$\displaystyle \frac{\partial F}{\partial x} = \frac{\partial }{\partial x} (\int_{x_{0}}^{x} f(t+iy) \; \mathrm{d}t) = f(x + iy) = f(z)$
$\therefore$ $\dfrac{\partial F}{\partial x} = -i \dfrac{\partial F}{\partial y}$
By Thm 2,
$f$ is continuous on $D$
Since $f$ is continuous function and Cor 1,
$F$ is differentiable at $z$
$\therefore$ $F$ is differentiable on $D$ and $F'(z) = f(z)$
$\therefore$ $F$ is primitive function of $f$
Cor 1
Let $D = D(z_{0}, r)$ and $f : D \to \mathbb{C}$ be analytic function.
Let $\Gamma : [a, b] \to D$ be a closed curve. Then
$\displaystyle \int_{\Gamma} f(z) \; \mathrm{d}z = 0$
Cor 2
Let $\Omega$ be a simply connected area and $f : \Omega \to \mathbb{C}$ be analytic function.
Let $\Gamma : [a, b] \to \Omega$ be a closed curve. Then
$\displaystyle \int_{\Gamma} f(z) \; \mathrm{d}z = 0$
Prop 1
$\displaystyle \int_{0}^{\infty} e^{-it^{2}} \; \mathrm{d}t = \dfrac{1}{2} \sqrt{\dfrac{\pi}{2}} (1 - i)$
Let $f(z) = e^{-z^{2}}$ . For $a >0$,
$\Gamma_{1}(t) = t$ $0 \leq t \leq a$
$\Gamma_{2}(t) = a + it$ $0 \leq t \leq a$
$\Gamma_{3}(t) = \dfrac{1}{\sqrt{2}}(1+i)t$ $0 \leq t \leq a \sqrt{2}$
By Cor 2,
$\displaystyle \int_{\Gamma_{1}} f(z) \; \mathrm{d}z + \int_{\Gamma_{2}} f(z) \; \mathrm{d}z = \int_{\Gamma_{3}} f(z) \; \mathrm{d}z$
Claim 1 : $\displaystyle \lim_{a \to \infty} \int_{\Gamma_{1}} f(z) \; \mathrm{d}z = \dfrac{\sqrt{\pi}}{2}$
Since $\displaystyle \int_{\Gamma_{1}} f(z) \; \mathrm{d}z = \int_{0}^{a} e^{-t^{2}} \; \mathrm{d}t$,
$\displaystyle \lim_{a \to \infty} \int_{\Gamma_{1}} f(z) \; \mathrm{d}z = \int_{0}^{\infty} e^{-t^{2}} \; \mathrm{d}t = \dfrac{\sqrt{\pi}}{2}$
end
Claim 2 : $\displaystyle \lim_{a \to \infty} \int_{\Gamma_{2}} f(z) \; \mathrm{d}z = 0$
Since $\left|e^{-(a+it)^{2}} \right| = e^{-a^{2} + t^{2}} \leq e^{-a^{2} + at}$ on $\Gamma_{2}$,
$\left|\displaystyle \int_{\Gamma_{2}} f(z) \; \mathrm{d}z \right| = \left|\displaystyle i \int_{0}^{a} e^{-(a + it)^{2}} \; \mathrm{d}t \right|$
$\leq e^{-a^{2}} \displaystyle \int_{0}^{a} e^{at} \; \mathrm{d}t$
$= \dfrac{1}{a} (1 - e^{-a^{2}})$
$\leq \dfrac{1}{a} $
$\therefore$ $\displaystyle \lim_{a \to \infty} \left|\displaystyle \int_{\Gamma_{2}} f(z) \; \mathrm{d}z \right| = 0$
$\therefore$ $\displaystyle \lim_{a \to \infty} \displaystyle \int_{\Gamma_{2}} f(z) \; \mathrm{d}z = 0$
end
Since $\displaystyle \int_{\Gamma_{3}} f(z) \; \mathrm{d}z = \dfrac{1}{\sqrt{2}} (1 + i) \int_{0}^{a \sqrt{2}} e^{-it^{2}} \; \mathrm{d}t$,
$\displaystyle \int_{0}^{\infty} e^{-it^{2}} \; \mathrm{d}t = \dfrac{\sqrt{2}}{1 + i} \dfrac{\sqrt{\pi}}{2} = \dfrac{1}{2} \sqrt{\dfrac{\pi}{2}} (1 - i)$
Cor 3
$\displaystyle \int_{0}^{\infty} \cos t^{2} \; \mathrm{d}t = \int_{0}^{\infty} \sin t^{2} \; \mathrm{d}t = \dfrac{1}{2} \sqrt{\dfrac{\pi}{2}}$
Thm 3
Let $\Omega$ be a simply connected area and $f : \Omega \setminus \left\{z_{0} \right\} \to \mathbb{C}$ be analytic function.
Let $\Gamma : [a, b] \to \Omega$ be a simple closed curve.
Let $z_{0}$ be inside $\Gamma$, and choose $r$ such that the circle $\left|z - z_{0} \right| = r$ is also inside $\Gamma$. Then
$\displaystyle \int_{\Gamma} f(z) \; \mathrm{d}z = \int_{\left|z - z_{0} \right| = r} f(z) \; \mathrm{d}z$
Let the curve of the circle $\left|z - z_{0} \right| = r $ be denoted as $C$
Draw a line passing through the center $z_{0}$ of the circle,
and let the points where this line intersects the circle be $A$ and $B$, respectively.
Let $D$ be the point where the ray starting from $z_{0}$ in the direction of $A$ meets the curve $\Gamma$,
and let $E$ be the point where the ray starting from $z_{0}$ in the direction of $B$ meets the curve $\Gamma$
Let the portion of the curve $\Gamma$ from $D$ to $E$ be denoted as $\Gamma_{1}$,
and the portion from $E$ to $D$ be denoted as $\Gamma_{2}$
Let the curve along the circle from $A$ to $B$ be denoted as $C_{1}$, and the curve from $B$ to $A$ as $C_{2}$
Let the line segment from $A$ to $D$ be denoted as $L_{1}$, and the line segment from $B$ to $E$ as $L_{2}$
Therefore
$L_{1} + \Gamma_{1} + (-L_{2}) + (-C_{1})$, $(-L_{1}) + (-C_{2}) + L_{2} + \Gamma_{2}$
are closed curves and function $f(z)$ is analytic inside them.
By Cor 2,
$\displaystyle \int_{L_{1} + \Gamma_{1} + (-L_{2}) + (-C_{1})} f(z) \; \mathrm{d}z = 0$
$\displaystyle \int_{(-L_{1}) + (-C_{2}) + L_{2} + \Gamma_{2}} f(z) \; \mathrm{d}z = 0$
By Prop 3,
$0 = \displaystyle \int_{L_{1} + \Gamma_{1} + (-L_{2}) + (-C_{1})} f + \displaystyle \int_{(-L_{1}) + (-C_{2}) + L_{2} + \Gamma_{2}} f$
$= \displaystyle \int_{L_{1}} f + \int_{\Gamma_{1}} f - \int_{L_{2}} f - \int_{C_{1}} f - \int_{L_{1}} f - \int_{C_{2}} f + \int_{L_{2}} f + \int_{\Gamma_{2}} f$
$= \displaystyle \int_{\Gamma_{1} + \Gamma_{2}} f - \int_{C_{1} + C_{2}} f $
$= \displaystyle \int_{\Gamma} f - \int_{C} f $
$\therefore$ $\displaystyle \int_{\Gamma} f(z) \; \mathrm{d}z = \int_{\left|z - z_{0} \right| = r} f(z) \; \mathrm{d}z$
Cor 4
Let $\Omega$ be an area and $f : \Omega \setminus \left\{z_{0} \right\} \to \mathbb{C}$ be analytic function.
Then for $\overline{D(z_{0}, r)} \subset \Omega$,
$\displaystyle \int_{\left|z -z_{0} \right| = r} f(z) \; \mathrm{d}z = \int_{\left|z - z_{0} \right| = \varepsilon} f(z) \; \mathrm{d}z$
( $\varepsilon \in (0, r)$ )
Since $\overline{D(z_{0}, r)} \subset \Omega$ and $\Omega$ is open, there exists $R > r$ s.t.
$D(z_{0}, R) \subset \Omega$
Since $D(z_{0}, R)$ is simply connected area, by Thm 3
$\displaystyle \int_{\left|z -z_{0} \right| = r} f(z) \; \mathrm{d}z = \int_{\left|z - z_{0} \right| = \varepsilon} f(z) \; \mathrm{d}z$