Complex Sequence and Series

sequence

sequence convergence

• $z_n=x_n+i{y}_n,z=x+iy\Rightarrow$

  $$\underset{n\to \infty }{\lim }{z}_n=z\iff \underset{n\to \infty }{\lim }{x}_n=x\text{ and }\underset{n\to \infty }{\lim }{y}_n=y$$

series

series convergence

$\sum z_n$ converge to $S$ $\iff$ partial sum $S_N$ converge to $S$

• $z_n=x_n+i{y}_n,S=X+iY\Rightarrow$

  $$\sum z_n=S\iff \sum x_n=X\text{ and }\sum y_n=Y$$

• divergence test

• absolute convergence

remainder

$${\rho }_N=S-S_N$$

• $\sum z_n$ converge $\iff {\lim }_{N\to \infty }{\rho }_N=0$

power series

see also power series

• ${\sum }_{n=0}^{\infty }{c}_n(z-z_0{)}^n$ converge to $S(z)$ for $z=z_1\ne z_0$
  $\Rightarrow$ absolutely convergent in circle $|z-z_0|<|z_1-z_0|$
  • circle of convergence, biggest such circle
• $S(z)$ is continuous function inside circle of convergence

power series integration

can integrate term by term within circle of convergence

$${\int }_{C}g(z)S(z)dz=\sum _{n=0}^{\infty }{a}_n{\int }_{C}g(z)(z-z_0{)}^{n}dz$$

• $S(z)$ holomorphic

  proof

  let $C$ be any closed contour and set $g(z)=1$

  $${\int }_{C}S(z)dz=\sum _{n=0}^{\infty }{a}_n{\int }_C(z-z_0{)}^{n}dz=\sum _{n=0}^{\infty }0=0$$

  by Morera’s theorem, $S(z)$ is holomorphic

• can differentiate $S(z)$ term by term

Taylor’s theorem

$f$ holomorphic in $|z-z_0|<R_0$

$\Rightarrow f$ analytic,

$$f(z)=\sum _{n=0}^{\infty }{a}_n(z-z_0{)}^n,a_n=\frac{f^{(n)}(z_0)}{n!}$$

uniqueness of Taylor series

in $|z-z_0|<R$, power series ${\sum }_{n=0}^{\infty }{a}_n(z-z_0{)}^n$ converge to $f(z)$

$\Rightarrow$ it is the Taylor series of $f$ about $z_0$

Last edited: 2024-09-13