This tutorial will be on the method of contour integration. In order to really appreciate this, one needs to study complex analysis. Since I wish to make this understandable to as many people as possible I will barely use any complex analysis. However, eventhough this is kept as simple as possible the reader needs to have some experience in math. The following are a must:

- know what complex numbers, do arithmetic with them, and geometrically represent them in the plane
- know basic Calculus such as differenciation and integration;
- know the basics of infinite series, such as computing radius of convergence.

The first thing we need to study are sequences and series of complex numbers. It is the exact same concept as with real sequences. A sequence $\displaystyle c_1,c_2,c_3,...$ of complex numbers is a function on the positive integers $\displaystyle \{ c_n \}$. We say $\displaystyle \lim ~ c_n = c$ when $\displaystyle c_n$ converges to the complex number $\displaystyle c$. Note, any complex number $\displaystyle z$ can be expressed in the form $\displaystyle x+iy$. Thus, we can think of $\displaystyle c_n = a_n + i b_n$ where $\displaystyle a_n$ and $\displaystyle b_n$ arerealsequences. This simplifies the problem of studing complex sequences into real sequences.

In fact, we have a simple theorem.

Theorem 1:Let $\displaystyle c_n = a_n + ib_n$ the sequence converges to $\displaystyle c=a+bi$ if and only if $\displaystyle \lim ~ a_n = a$ and $\displaystyle \lim ~ b_n = b$.

Example 1:Consider $\displaystyle c_n = \frac{i^n}{n}$. Write out the first terms to see what is going on $\displaystyle \frac{i}{1}, \frac{-1}{2}, \frac{-i}{3}, \frac{1}{4}, ... $. We see the pattern that $\displaystyle a_n = 0$ if $\displaystyle n$ is odd and $\displaystyle a_n = \frac{(-1)^{n/2}}{n}$ if $\displaystyle n$ is even. Thus, the first few terms for $\displaystyle a_n = 0, -\frac{1}{2}, \frac{1}{4},...$. Similarly, the first few terms for $\displaystyle b_n$ are $\displaystyle 1,-\frac{1}{3},\frac{1}{5},...$. Note, $\displaystyle \lim ~ a_n = \lim ~ b_n = 0$. Thus, $\displaystyle \lim ~ c_n = 0 + 0i = 0$.

Now we get to another important concept, series. It is assumed that the reader knows some series from Calculus, otherwise this will be hard. Let $\displaystyle c_1,c_2,...$ be a complex-valued sequence. Define $\displaystyle s_n = \sum_{k=1}^n c_k$. Thus, $\displaystyle s_1 = c_1, s_2 = c_1+c_2, s_3 = c_1+c_2+c_3, ... $. This is called thesequence of partial sums. We say $\displaystyle \sum_{n=1}^{\infty} c_n = c$ for a complex number $\displaystyle c$ iff $\displaystyle \lim ~ s_n = c$. We say $\displaystyle \{ c_n \}$ is aconvergent seriesin that case. Otherwise, we saydivergent.

In what follows the absolute value $\displaystyle |z|$ of a complex number $\displaystyle z=x+iy$ is defined to be $\displaystyle \sqrt{x^2+y^2}$. Thus, the absolute value of a complex number is always a non-negative real number.

Example 2:Just like in Calculus we have an analouge of a geometric series. Let $\displaystyle |z| < 1$. Let us prove that $\displaystyle \sum_{n=0}^{\infty} z^n $ is convergent, and furthermore find its sum. Note that, $\displaystyle 1+z+z^2+...+z^n = \frac{1-z^n}{1-z}$. When we take the limit it turns out that $\displaystyle \lim ~ z^n = 0$ (when $\displaystyle |z|<1$). But the problem with proving this is that there is no nice way of writing $\displaystyle (x+iy)^n$, if we want to seperate the real and imaginary parts like in Example 1. So we will prove this in a nice way latter on. But if we accept $\displaystyle \lim ~ |z|^n = 0$ then it means $\displaystyle \lim ~ \frac{1-z^n}{1-z} = \frac{1}{1-z}$. Thus, $\displaystyle \sum_{n=0}^{\infty}z^n = \frac{1}{1-z}$.

In Calculus we study power series. It is a function $\displaystyle f(x) = \sum_{n=0}^{\infty} a_n x^n$, the domain of the function are all values $\displaystyle x$ where it converges. The question is of course how to find where it converges. We usually use the ratio test, by writing, $\displaystyle \lim~ \left| \frac{a_{n+1}x^{n+1}}{a_n x^n} \right| = \lim ~ \left| \frac{a_{n+1}}{a_n} \right||x| < 1$. It needs to be less than 1 be have (absolute) convergence. When it is larger than 1 it diverges. Note, if the limit fails to exists (not even infinity) then the test fails. Or when sum of the $\displaystyle a_n$ are zero the test fails also because then we divide by zero. There is also the ratio test by taking the limit $\displaystyle \lim ~ |a_nx^n |^{1/n} = \lim ~ |a_n|^{1/n} |x| < 1$, again if the limit is larger than 1 then it diverges. When working with the root test an important limit to know is that $\displaystyle \lim~ n^{1/n} = 1$. In complex analysis, the exact same thing happens. In fact, you can write out the summation in terms of real and imaginaries to prove this.

Example 3:Find the radius of convergence of $\displaystyle \sum_{n=0}^{\infty} z^n$. Let us use the root test and we get $\displaystyle \lim ~ |z^n|^{1/n} = \lim ~ |z| = |z|$. Thus, if $\displaystyle |z| < 1$ we have convergence. If $\displaystyle |z| > 1$ we have divergence. But just like in Calculus we need to check the endpoints, but unlike in Calculus where there are just two points here there are infinitely many! All lieing on $\displaystyle |z| = 1$, i.e. the circle $\displaystyle x^2 + y^2 = 1$. This problem turns out being not as simple, so we will simply ignore it because we will not need it (in case you are interested apply the divergence test and claim the series will diverge at $\displaystyle |z|=1$). However, the radius of convergence is $\displaystyle R=1$.

Note in complex analysis the termradius of convergencemakes more sense than in Calculus. Because the radius of convergence is the circle on which the power series converges. Remember we say $\displaystyle R=0$ if the power series does not convergence anywhere, which is not interesting and we will never have that. And $\displaystyle R=\infty$ if the power series converges everywhere in complex plane, i.e. a circle of infinite radius of convergence.

Example 4:Consider $\displaystyle f(z) = \sum_{n=0}^{\infty}\frac{z^n}{n!}$. Here it is easier to use ratio test $\displaystyle \lim ~ \left| \frac{z^{n+1}}{(n+1)!}\cdot \frac{n!}{z^n} \right| = \lim~ \frac{|z|}{n+1} = 0$ because $\displaystyle n+1 \to \infty$ while $\displaystyle |z|$ stays fixed. Thus for any $\displaystyle z$ in the complex plane the ratio limit is strictly less than 1, thus we have convergence. Since this converges everywhere we have $\displaystyle R=\infty$.

Example 5:Consider $\displaystyle f(z) = \sum_{n=0}^{\infty} \frac{2^nz^{2n}}{n^{100}}$. Here is easier to use root test. $\displaystyle \lim ~ \left| \frac{2^nz^{2n}}{n^{100}} \right|^{1/n} = \lim ~ \left| \frac{2z^2}{(n^{1/n})^{100}} \right|= 2|z|^2 < 1$. Note, we do not have $\displaystyle |z|^2 = z^2$, that is clearly wrong since $\displaystyle z$ is a complex variable so the square can be negative. Thus, $\displaystyle |z|^2 < \frac{1}{2} \implies -\frac{1}{\sqrt{2}} < |z| < \frac{1}{\sqrt{2}}$. Since $\displaystyle |z|\geq 0$ it is equivalent to simply writing $\displaystyle |z| < \frac{1}{\sqrt{2}}$. Thus, the radius of convergence is $\displaystyle R=\frac{1}{\sqrt{2}}$.