Key points to keep in mind:
Triangle inequalities: \big| |a|-|b| \big| \le |a+b| \le |a| + |b| Exact values of trigonometric functions: \begin{aligned} \sin 0&= 0&\cos 0&=1 &\tan 0 &= 0\\ \sin \pi/6&= 1/2 &\cos \pi/6&=\sqrt 3/2 &\tan \pi/6&=1/\sqrt3 \\ \sin \pi/4&= \sqrt 2/2 &\cos \pi/4&=\sqrt 2/2 &\tan \pi/4&=1\\ \sin \pi/3&= \sqrt 3/2&\cos \pi/3&=1/2 &\tan \pi/3&=\sqrt 3 \\ \sin \pi/2&= 1&\cos \pi/2&=0 &\tan \pi/2&\quad \text{undef.} \end{aligned}
A mapping of the form T(z) = w = ({az+b})/({cz+d}).
Has inverse T^{-1}(w) = (-dw+b)/(cw-a).
Can be extended to \overline{\mathbb C} with \begin{aligned} T(\infty) = a/c &\iff \lim_{|z|\to\infty} T(z) = a/c\\ T(-d/c) = \infty &\iff \lim_{z\to -d/c} 1/T(z) = 0 \end{aligned}
\begin{aligned} \cos z &= \frac{e^{iz}+e^{-iz}}2&\sin z &= \frac{e^{iz}-e^{-iz}}{2i} \\ \cosh x &= \frac{e^z + e^{-z}}2& \sinh x &= \frac{e^z-e^{-z}}2 \end{aligned}
\begin{aligned} \lim_{z \to \infty} f(z) &= w_0 &&\iff &\lim_{z \to 0} f(1/z) &= w_0 \\ \lim_{z \to z_0} f(z) &= \infty &&\iff &\lim_{z \to z_0} \frac 1 {f(z)}& = 0\\ \lim_{z \to \infty} f(z) &= \infty&& \iff& \lim_{z \to 0} \frac 1{f(1/z)} &= 0 \end{aligned}
\frac{\partial}{\partial z}=\frac 1 2\left( \frac {\partial}{\partial x}-i\frac {\partial}{\partial y} \right)\quad\text{and}\quad \frac{\partial}{\partial \bar z}=\frac 1 2\left( \frac {\partial}{\partial x}+i\frac {\partial}{\partial y} \right).
The following are equivalent:
Suppose f is continuous on a contour C, given by z = z(t) and a \le t \le b. Then, there exists M such that |f(z)| \le M for all z \in C and \left|\int_C f(z)\,dz\right| \le M \int_a^b |z'(t)|\,dt = M\ell where \ell is the arc length of C.
If f is analytic on a contour C, as well as on C_1, \ldots, C_n \subset \operatorname{Int}C and on the interior of the domain bordered by C_1, C_2, \ldots, C_n, and C, C_1, \ldots, C_n are all positively oriented, then \int_C f(z)\,dz + \sum_{j=1}^n \int_{C_j}f(z)\,dz = 0. Positively oriented means the that while traversing the contour, the region is on your left. This is particularly important for the orientation of C_1, \ldots, C_n.
Let f be analytic on and inside a simple closed curve C, positively oriented. Then, if z_0 \in \operatorname{Int}C we have \begin{aligned} f(z_0) &= \frac 1 {2\pi i} \int_C \frac{f(z)}{z-z_0}\,dz. \end{aligned} Also under the same conditions, f^{(n)}(z_0) = \frac{n!}{2\pi i}\int_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz.
The Dirichlet boundary problem \begin{cases} \Delta u = 0 & \text{in }B_{r_0}, \\ u|_{\partial B_{r_0}} = \Phi(r_0, \phi)& \text{on }\partial B_{r_0}, \end{cases} is solved by u(r, \theta) = \frac 1 {2\pi}\int_0^{2\pi}\underbrace{\frac{r_0^2-r^2}{r_0^2-2r_0r\cos(\phi-\theta)+r^2}}_{P(r_0,r,\phi,\theta)}\Phi(r, \phi)\,d\phi.
To calculate radius of convergence of a power series \sum a_n (z-z_0)^n, we can use the ratio test by computing \Lambda = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| \quad \implies \quad R = \frac 1 \Lambda. Conventionally, \Lambda=0 \iff R= \infty and \Lambda = \infty \iff R=0.
Some common Taylor series are below. \begin{aligned} \exp z &= \sum_{n=0}^\infty \frac {z^n}{n!} \\ \frac {1}{1-z} &= \sum_{n=0}^\infty z^n \quad \text{for }|z|<1 \\ \sin z&= \sum_{n=0}^\infty \frac {(-1)^nz^{2n+1}}{(2n+1)!} \\ \sinh z&= \sum_{n=0}^\infty \frac {z^{2n+1}}{(2n+1)!} \end{aligned}
Laurent series are defined in an annulus centred at some point (not necessarily a singularity) and exist in a radius around that point to the first singularity. To find Laurent series expansions in regions of the form \{z : |z| > r\}, we can manipulate the series into the form 1/(1-r/z) and use the geometric series.
The residue is the coefficient of the z^{-1} term of a Laurent series. If z_0 is an isolated singularity of f, then \operatorname*{Res}_{z=z_0}f(z)=\frac 1 {2\pi i}\int_C f(\xi)\,d\xi.
Suppose C is a positively oriented simple closed curve and that f is analytic in and on C except at finitely many isolated points \{z_1, z_2, \ldots, z_k\}. Then, \int_C f(z)\,dz = 2\pi i \sum_{j=1}^k \underset{z=z_j}{\operatorname{Res}}f(z).
Assume the Laurent series on B_R(z_0) \setminus \{z_0\} is given by f(z) = \sum_{n=0}^\infty a_n(z-z_0)^n + \sum_{n=1}^\infty b_n(z-z_0)^{-n}.
This can be used to easily compute residues: f has a pole of order N at z_0 if and only if we can write f(z) = \frac{\phi(z)}{(z-z_0)^N} inside B_r(z_0)\setminus \{z_0\} where \phi is analytic in B_r and \phi(z_0)\ne 0. Moreover, \operatorname*{Res}_{z=z_0} f(z) = \frac{\phi^{(N-1)}(z_0)}{(N-1)!}.
f is analytic at z_0 and has a zero of order m at z_0 if and only if f(z) = (z-z_0)^m g(z) where is analytic and g(z_0)\ne 0.
Suppose p and q are analytic at z_0, p(z_0)\ne 0, and q has a zero of order m at z_0. Then, p/q has a pole of order m at z_0.
Let p, q be analytic at z_0. If p(z_0)\ne 0 and q(z_0)= 0, then p/q has a simple pole at z_0 and \operatorname{Res}_{z=z_0} p /q = {p(z_0)}/{q'(z_0)}.
Defined as \operatorname*{PV}\int_{-\infty}^\infty f = \lim_{m\to\infty}\int_{-m}^mf(x)\,dx.
When the PV coincides with the regular integral (e.g. for even or non-negative functions), \int_0^\infty f(x)\,dx=\frac 1 2 \int_{-\infty}^\infty f(x)\,dx = \frac 1 2 \operatorname{PV} \int_{-\infty}^\infty f(x)\,dx.
Let f and g be analytic in and on a simple closed curve C (orientation irrelevant). Suppose |g(z)| < |f(z)| for all z on this curve. Then, f and f+g have the same number of zeros (counting multiplicity) inside C.