Complex Analysis Domain Guide

Complex differentiability requires the same limit from every direction in the plane, and that is an enormously restrictive demand. The Cauchy–Riemann equations are exactly the price: the real and imaginary parts cannot vary independently. Everything that follows — infinite differentiability, power-series representation, the maximum modulus principle, Liouville's theorem — is the payoff for a hypothesis that is much stronger than it looks.

Integration in the complex plane is topological. Cauchy's theorem says a loop enclosing nothing singular contributes nothing, so a contour integral depends only on which singularities the loop encircles. That converts integration into bookkeeping: find the singularities inside, add up their residues, multiply by $2\pi i$.

The residue is the one Laurent coefficient that does not integrate away. Every term $(z-a)^n$ with $n \ne -1$ has a single-valued antiderivative around a loop and contributes zero; $(z-a)^{-1}$ has $\log$ as its antiderivative, which changes by $2\pi i$ each circuit. That is the whole content of the residue theorem in one sentence.

Complex arithmetic first, because several items each year need nothing else. In polar form $z = re^{i\theta}$, multiplication multiplies moduli and adds arguments, so de Moivre's theorem $z^n = r^n e^{in\theta}$ is immediate. The $n$-th roots of a nonzero $w = \rho e^{i\phi}$ are $\rho^{1/n}e^{i(\phi + 2\pi k)/n}$ for $k = 0, \dots, n-1$ — equally spaced on a circle. The $n$-th roots of unity are the case $w = 1$; they form a cyclic group of order $n$ under multiplication, and for $n \ge 2$ they sum to zero, since they are the roots of $z^n - 1$ whose next-highest coefficient vanishes. Euler's identity $e^{i\theta} = \cos\theta + i\sin\theta$ also gives $\cos\theta = (e^{i\theta} + e^{-i\theta})/2$ and $\sin\theta = (e^{i\theta} - e^{-i\theta})/(2i)$, which is why $\sin z$ and $\cos z$ are unbounded on $\mathbb{C}$ even though their real restrictions are not.

Analyticity is tested with the Cauchy–Riemann equations. For $f = u + iv$, differentiability requires $u_x = v_y$ and $u_y = -v_x$, and those conditions plus continuity of the partials are sufficient. When they hold, $f'(z) = u_x + iv_x$. The equations immediately explain why $\bar{z}$, $|z|$, and $\operatorname{Re}(z)$ are nowhere analytic, and why $u$ and $v$ must each be harmonic: differentiating the two equations and adding gives $u_{xx} + u_{yy} = 0$. Constructing a harmonic conjugate — given harmonic $u$, find $v$ with $f = u + iv$ analytic — is a standard exam task solved by integrating $v_y = u_x$ and then fixing the arbitrary function of $x$ using $v_x = -u_y$.

Integration begins with Cauchy's theorem: if $f$ is analytic on and inside a simple closed contour, the integral around it is zero. Cauchy's integral formula extends this to a function with one non-analytic factor: $f(a) = \dfrac{1}{2\pi i}\displaystyle\oint_C \frac{f(z)}{z - a}\,dz$ for $a$ inside $C$, and differentiating under the integral gives $f^{(n)}(a) = \dfrac{n!}{2\pi i}\displaystyle\oint_C \frac{f(z)}{(z-a)^{n+1}}\,dz$. That formula is the source of the rigidity results: it shows an analytic function has derivatives of all orders, and the resulting Cauchy estimates give Liouville's theorem — a bounded entire function is constant — from which the fundamental theorem of algebra follows in a few lines.

When analyticity fails at isolated points, expand in a Laurent series on an annulus. The principal part classifies the singularity: no negative powers means removable (the function extends analytically, as $\sin z / z$ does at $0$); finitely many, stopping at $(z-a)^{-m}$, means a pole of order $m$; infinitely many means essential, as $e^{1/z}$ has at $0$. The classification can also be made by limits: the singularity is removable when $\lim_{z \to a} f(z)$ is finite, a pole when $|f(z)| \to \infty$, and essential otherwise — near an essential singularity the function comes arbitrarily close to every complex value, which is why no limit exists.

Residues turn all of this into computation. At a simple pole, $\operatorname{Res}_{z=a} f = \lim_{z \to a}(z-a)f(z)$, and when $f = p/q$ with $q$ having a simple zero at $a$, this is just $p(a)/q'(a)$. At a pole of order $m$, $\operatorname{Res}_{z=a} f = \dfrac{1}{(m-1)!}\lim_{z\to a}\dfrac{d^{m-1}}{dz^{m-1}}\left[(z-a)^m f(z)\right]$. The residue theorem then states that for a positively oriented simple closed contour enclosing finitely many isolated singularities, $\oint_C f = 2\pi i \sum \operatorname{Res}$. Essential singularities have residues too, but they must be read off the Laurent series rather than computed by a derivative formula — for $e^{1/z}$ at $0$ the expansion $1 + z^{-1} + \tfrac{1}{2}z^{-2} + \cdots$ gives residue $1$.

The standard application is a real improper integral. To evaluate $\int_{-\infty}^{\infty} R(x)\,dx$ for a rational $R$ whose denominator has no real zeros and whose degree exceeds the numerator's by at least $2$, integrate $R(z)$ over the boundary of a half-disc of radius $R$ in the upper half-plane. The residue theorem gives the value of the closed contour integral; an ML estimate shows the semicircular arc contributes something of size $O(1/R)$, which vanishes; so the real integral equals $2\pi i$ times the sum of the residues in the upper half-plane. The same scheme with a factor $e^{iaz}$ handles integrals of $R(x)\cos ax$ and $R(x)\sin ax$, where Jordan's lemma supplies the decay.