🧔‍♀️ Theorem: If $f(z)$ is analytic in and on a contour $C$ which encloses the point $z=a$ then

$$ f(a)=\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-a)}\,\text dz $$

💫 Proof: Consider circle $C$ of radius $\epsilon$ centred on $a$ so that $z=a+\epsilon e^{i\theta}$ and $\text dz=i\epsilon e^{i\theta}\,\text d\theta$

$$ \begin{aligned} \oint \frac{f(z)}{z-a}\,\text dz&=\int ^{2\pi}_0 \frac{f(a+\epsilon e^{i\theta} )}{\epsilon e^{i\theta}} i\epsilon e^{i\theta}\, \text d\theta \\ &=i\int_0^{2\pi} f(a+\epsilon e^{i\theta} ) \, \text d\theta \end{aligned} $$

Valid for any $\epsilon >0$

• We reduce $\epsilon$ to the point

  $$ \begin{aligned} \oint \frac{f(z)}{z-a}\,\text dz&=i\lim _{\epsilon \to 0} \int ^{2\pi}_0 f(a+\epsilon e^{i\theta})\,\text d\theta \\ &=i\int^{2\pi}_0 f(a)\, \text d \theta= 2\pi i f(a) \\ \therefore f(a)&=\frac{1}{2\pi i} \oint_C \frac{f(z)}{(z-a)}\,\text dz

  \end{aligned} $$

🗒️ Note: this also works for the derivative of $f(z)$

$$ f'(a)=\frac{1}{2\pi i} \oint _C \frac{f(z)}{(z-a)^2}\,\text dz $$

• Proof:

<aside> 🎚️ General Cauchy’s integral to any derivative order of $n$

$$ \frac{\text d^n f(a)}{\text da^n}=f^{(n)} (a)=\frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z-a)^{n+1}}\text dz $$

as $0!=1$, this includes the original Cauchy integral formula

</aside>

💃 Example: Consider the integral $I=\oint_C \frac{\cos(z)}{z}\,\text dz$ for some contour $C$ that encloses $z=0$

• Applying Cauchy’s integral with $a=0$ such that $\frac{1}{z-a}=\frac{1}{z}$ and $f(z)=\cos z$ we have

  $$ \begin{aligned} I&=2\pi i f(a=0) \\ &=2\pi i \cos(0)= 2\pi i \end{aligned} $$

Consequences

1. $f^{(n)}(a)$ expressed in terms of $f(z)$ on $C$ so all derivatives exist

2. If $f(z)$ is analytic in a region $R$ then $F(z)=\int^z_a f(\xi)\,\text d \xi$ is path independent, $\frac{\text dF}{\text dz}=f(z)$ and thus $F(z)$ is analytic.

3. Morea’s theorem: If $\oint_C f(z)\,\text dz = 0$ for any path $C$ within a region $R$, then $f(z)$ is analytic within $R$.

4. Liouville’s theorem: Every function that is bounded and entire must be constant

5. Cauchy’s inequality: For $f(z)$ analytic within and on a circle $C$ of radius $R$ then:

   $$  \left| f^{(n)}(a) \right| \leq \frac{M n!}{R^n} $$

   where $M$ is the maximum modulus of $f(z)$ on $C$

6. Fundamental Theorem of Algebra: If a polynomial $P(z)$ has no roots, $\frac{1}{P(z)}$ is entire and bounded, contradicting the assumption that $P(z)$ is a polynomial

Argument theorem

🧔‍♀️ Theorem: If $f(z)$ is analytic except at $P$ poles and has $N$ zeros within some curve $C$

$$ \frac{1}{2\pi i}\oint_C\frac{f'(z)}{f(z)}\,\text dz=N-P $$

💫 Proof: Start with a function with one pole of order $p$ and one zero of order $n$ within $C$

$$ \begin{aligned} f(z)=g(z)\frac{(z-\beta)^n}{(z-\alpha)^p} \end{aligned} $$

where $g(z)$ is analytic with no zeros within $C$.

• The derivative of $f(z)$ is then

  $$ f'(z)=g'(z)\frac{(z-\beta)^n}{(z-\alpha)^p}+ng(z)\frac{(z-\beta)^{n-1}}{(z-\alpha)^p}-pg(z)\frac{(z-\beta)^n}{(z-\alpha)^{p+1}} $$

  So

  $$ \frac{f'(z)}{f(z)}=\frac{g'(z)}{g(z)}+\frac{n}{z-\beta}-\frac{p}{z-\alpha} $$

• Solving the integrtal

  $$ \oint_C\frac{f'(z)}{f(z)}\,\text dz=\underbrace{\oint_C\frac{g'(z)}{g(z)} \,\text dz}_{=0 \text{ no zero within C}}+2\pi i(n-p)=2\pi i(n-p) $$

This can be replicated for more zeros and poles and we get $N=\sum{n_i}$ and $P=\sum p_i$