🧔‍♀️ Theorem: If $f(z)$ is analytic within a circle of radius $R$ centered on $a$ then for all $z$ such that $|z-a|<R$ we can express $f(z)$ as a power series

$$ f(z)=f(a)+f'(a)(z-a)+\frac{1}{2!}f''(a)(z-a)^2+\ldots +\frac{1}{n!}f^{(n)}(z-a)^n+\ldots $$

💫 Proof: Let $C$ be a circular path within a region of convergence of $f(z)$ and let $z$ and $a$ be within this path

• Using Cauchy’s integral:

$$ \begin{aligned} f(z)&=\frac{1}{2\pi i}\oint_C \frac{f(\xi)}{\xi-z} \,\text d \xi \\ &=\frac{1}{2\pi i}\oint_C \frac{f(\xi)}{(\xi-a)-(z-a)}\,\text d\xi \end{aligned} $$

Making $|\xi-a|>|z-a|$ we get

$$ \begin{aligned} f(z)&= \frac{1}{2\pi i} \oint C \frac{f(\xi)}{\xi-a} \left ( 1-\frac{z-a}{\xi-a} \right )^{-1} \\ &= \frac{1}{2\pi i} \oint_C \frac{f(\xi)}{\xi-a}\left ( 1+\frac{z-a}{\xi-a}+ \left ( \frac{z-a}{\xi-a} \right )^2+ \ldots \right ) \\ &=\sum^\infin{n=0} (z-a)^n \frac{1}{n!}f^{(n)}(a) \end{aligned} $$

💃 Example: $f(z)=(1-z)^{-1}$ about $z=1$

$$ \begin{aligned} f'(z)=-\frac{1}{(1+z)^2} \quad f''(z)=\frac{2}{1+z^3} \quad f'''(z)=-\frac{6}{(1+z)^4} \qquad f^{(n)}&= \frac{(-1)^n n!}{(1+z)^{n+1}}\\ f(1)=\frac{1}{2} \qquad f'(1)=-\frac{1}{4} \qquad f''(1)=\frac{1}{8} \qquad f^{(n)}&=\frac{(-1)^n n!}{2^{n+1}} \end{aligned} $$

• Combining we get

  $$ \begin{aligned} f(z)&= f(1)+f'(1)(z-1)+f''(1)(z-1)^2 + \ldots \\ &= \frac 12 - \frac 14 (z-1)+ \frac 18 (z-1)^2 \frac 1{16}(z-1)^3 + \ldots + \frac{(-1)^n}{2^{n+1}}(z-1)^n+ \ldots \end{aligned} $$

Laurent’s theorem

💼 Case: Consider a function $f(z)$ which is analytic for $R_1<|z-a|<R_2$, taken around two different contours $C_1$ and $C_2$

• $f(z)$ is analytic within $C$ so

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

  We can write this as the sum of 4 integrals

  $$ \oint_C=\int^B_A +\int^C_B +\int^D_C+\int ^A_D $$

• We can shrink the gap until

  $$ \int^A_D\to \oint _{C_1} \qquad \int ^C _B \to-\oint _{C_2} \qquad \int ^B_A \to -\int^D_C $$

• Allowing us to write

  $$ f(z)=\frac{1}{2\pi i} \oint_{C_1} \frac{f (\xi)}{\xi -z } \,\text d\xi-\frac{1}{2\pi i} \oint_{C_2} \frac{f (\xi)}{\xi -z } \,\text d\xi $$

• Starting with left hand expression

  $$ \begin{aligned} \frac{1}{\xi - z}&=\frac{1}{(\xi - a)-(z-a)}=\frac{1}{\xi - a} \left ( 1- \frac{z-a}{\xi -a} \right ) ^{-1} \\ &=\frac{1}{\xi-a} \left ( 1+\frac{z-a}{\xi-a}+\left ( \frac{z-a}{\xi-a} \right )^2+\ldots \right )

  \end{aligned} $$

  Thus we can expand the expression

  $$ \begin{aligned} \frac{1}{2\pi i } \oint _{C_1}\frac{f(\xi)}{\xi-z}\,\text d\xi&=\frac{1}{2\pi i } \oint _{C_1}\frac{f(\xi)}{\xi-a}\,\text d\xi+\frac{z-a}{2\pi i } \oint _{C_1}\frac{f(\xi)}{(\xi-a)^2}\,\text d\xi+\frac{(z-a)^2}{2\pi i } \oint _{C_1}\frac{f(\xi)}{(\xi-a)^3}\,\text d\xi \\ &\; +\ldots + \frac{(z-a)^n}{2\pi i } \oint {C_1}\frac{f(\xi)}{(\xi-a)^{n+1}}\,\text d\xi + \ldots \\ &= \sum^\infin{n=0} a_n (z-a)^n \\ \text{where} \quad a_n&=\frac{1}{2\pi i } \oint _{C_1}\frac{f(\xi)}{(\xi-a)^{n+1}}\,\text d\xi \end{aligned} $$

• The same can be done for the right hand side to get

  $$ \begin{aligned} \frac{1}{2\pi i } \oint {C_1}\frac{f(\xi)}{\xi-z}\,\text d\xi & = \sum^\infin{n=1} \frac{b_n}{(z-a)^n} \\ \text{where} \quad b_n&=\frac{1}{2\pi i } \oint _{C_2}f(\xi)(\xi-a)^{n-1}\,\text d\xi \end{aligned} $$

<aside> 🥫 Laurent series:

$$ f(z)=\sum_{n=0} ^\infin a_n(z-a)^n +\sum^\infin_{n=1} b_n \frac{1}{(z-a)^n} $$

</aside>