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$\newcommand{\R}{\mathbb{R}}$ $\newcommand{\RR}{\mathbb{R}}$ $\newcommand{\C}{\mathbb{C}}$ $\newcommand{\N}{\mathbb{N}}$ $\newcommand{\Z}{\mathbb{Z}}$
Complex derivatives
For a real function $f : \RR^2\to\RR^2$, we have that the total derivative is a linear map $T$ such that:
$$ f(a+h)  f(a)  T(h) = o_{lim_{h\to0}}(h) $$
with $\frac{\partial f_i(a)}{\partial x_j}=T_{ij}$.
For $g : \C \to \C$, differentiability is defined the same way, with the crucial difference that $T$ is linear in a vector space over a complex field, i.e. $T$ is the operator corresponding to multiplication by a complex number. Multiplication by a complex number corresponds to the following matrix:
$$ \begin{bmatrix} x & y & \\ y & x & \end{bmatrix} $$
(The reason for this is that a complex number corresponds to a stretch followed by a rotation; composition of this results in a matrix of the above form).
As a result,
$$ \frac{\partial f_1(a)}{\partial x_1} = \frac{\partial f_2(a)}{\partial x_2} $$
and
$$ \frac{\partial f_1(a)}{\partial x_2} = \frac{\partial f_2(a)}{\partial x_1} $$
These are the CauchyRiemann equations.
Complex functions
For a function $f$, a pole is a zero of $\frac{1}{f}$.
A holomorphic function is a complex function which is complex differentiable in a neighborhood of every point in its domain. Every holomorphic function is analytic: this is an essential difference to real analysis: roughly, if a complex function is once differentiable, it is infinitely differentiable.
A function whose only singularities are poles is meromorphic.
Contour integrals
These are line integrals of holomorphic functions around a loop in the complex plane. Cauchy’s integral theorem (which follows from Stokes’ theorem) states that these are invariant under homotopy of paths. Simplest example:
$$ \oint_C z^kdz$$
Because the domain is the whole real plane (no punctures), we can assume WLOG that the path is the unit circle $\gamma$ around the origin. Then suppose $\theta$ is defined such that $z=e^{i\theta}$, so that $dz = d(e^{i\theta})= ie^{i\theta}d\theta$. Then
$$ \oint_{\gamma} (e^{i\theta})^kie^{i\theta}d\theta = i\oint_{\gamma} e^{i(k+1)\theta}d\theta$$
This evaluates to $2\pi i$ if $k=1$ but otherwise to $0$. This fact lets us compute more complex contour integrals, like:
$$ \oint_{\gamma} \frac{e^z}{z^5}dz = \oint_{\gamma} \frac{\sum_{k=0}^{\infty}z^k}{z^5k!}dz = \oint_{\gamma} z^{1}4!dz = \frac{\pi i}{12} $$
In the above derivation, we expressed $e^z$ by its Taylor series, distributed the integral over the sum (which we can do because the function converges uniformly on the support of $\gamma$) and then noted that only the term $\frac{z^4}{z^54!}$ survives, by the above result.
In a very similar vein:
$ \oint_{\gamma} \frac{f(z)}{z^2}dz = \oint_{\gamma} \frac{f(0)+f’(0)z+\ldots}{z^2}dz = 2\pi if’(0) $
Residue
Note that $\oint_{\gamma} z^kdz$ is $0$ whenever $k\neq 1$. This is in line with Cauchy’s integral theorem, which allows us to contract any loop to a point as long as there is no singularity inside the loop, resulting in the integral equal to $0$.
The residue of a singularity is the result of an integral around just that singularity (up to a factor of $2\pi i$). It is important because the residue theorem allows for the calculation of a contour integral by (roughly) summing up the residues times their winding numbers (number of times the path loops around the singularity).
Using complex analysis to solve real integrals
It seems like a common trick is to take an integral on the real line, extend it to a loop going through the complex plane, and then using the powerful tools of complex analysis (e.g. residue theorem) to solve it.