Complex Analysis in a Nutshell

$$ \begin{array}{l} z \in \mathbb{C}, z = x + iy, x,y, \in \mathbb{R} \\ u(x,y), v(x,y) \in \mathbb{R}^2 \mapsto \mathbb{R} \\ f(z) = u(x,y) + iv(x,y), \ dz = dx + idy \end{array} $$

Cauchy-Riemann Condition (for analyticy)

$$ \begin{array}{l} f'(z) = \frac{f(z + dz) - f(z)}{z + dz - z} = \frac{f(z + dz) - f(z)}{dz} \\ = \frac{[ u(x+dx, y+dy) - u(x,y)] + i([v(x+dx,y+dy) - v(x,y)]}{dx + idy } \\ dx \to 0: f'(z) = -i \frac{\partial u}{\partial y} + \frac{\partial v}{\partial y} \\ dy \to 0: f'(z) = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \\ \to \frac{\partial v}{\partial y} - i \frac{\partial u}{\partial y} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \\ \boxed{ \frac{\partial v}{\partial y} = \frac{\partial u}{\partial x}, \ \ -\frac{\partial u}{\partial y} = \frac{ \partial v}{\partial x} } \\ \end{array} $$

Greens Theorem: $\Omega$ analytic

$$ \begin{array}{l} \boxed{ \int\limits _ {\partial \Omega} p\ dx = - \iint\limits _ {\Omega} p _ y\ dx\ dy } \end{array} $$

Proof:

$$ \begin{array}{llr} \int\limits _ {\partial \Omega} p\ dx & = \int\limits _ { \Gamma _ 0} p(x,y) dx + \int\limits _ {\Gamma _ 2} p(x,y) dx & \\ & = \int\limits _ {x _ 0}^{x _ 1} p(x, y _ 0) dx + \int\limits _ {x _ 1}^{x _ 0} p(x,y _ 1) dx & \\ & = - \int\limits _ {x _ 0} ^ {x _ 1} ( p(x,y _ 1) - p(x, y _ 0)) dx & \\ & = - \int\limits _ {x _ 0}^{x _ 1} \int\limits _ {y _ 0}^{y _ 1} p _ y \ dy \ dx & \\ & & \blacksquare \\ \end{array} $$

$$ \begin{array}{l} \boxed{ \int\limits _ {\partial \Omega} q\ dy = \iint\limits _ {\Omega} q _ x\ dx\ dy } \end{array} $$

Proof:

$$ \begin{array}{llr} \int\limits _ {\partial \Omega} q\ dy & = \int\limits _ { \Gamma _ 1} q(x,y) dy + \int\limits _ {\Gamma _ 3} q(x,y) dy & \\ & = \int\limits _ {y _ 0}^{y _ 1} q(x _ 1,y) dy + \int\limits _ {y _ 1}^{y _ 0} q(x _ 0, y) dy & \\ & = \int\limits _ {y _ 0} ^ {y _ 1} ( q(x _ 1, y) - q( x _ 0, y)) dy & \\ & = \int\limits _ {y _ 0}^{y _ 1} \int\limits _ {x _ 0}^{x _ 1} q _ y \ dx \ dy & \\ & & \blacksquare \\ \end{array} $$

Cauchy Theorem

$f(z)$ analytic in $\Omega$:

$$ \begin{array}{ll} \to & \boxed{ \oint\limits _ {\partial \Omega} f(z)\ dz = 0 } \\ \end{array} $$

Proof:

$$ \begin{array}{llr} \oint\limits _ {\partial \Omega} f(z)\ dz & = \oint\limits _ {\partial \Omega} (u + iv)(dx + idy) & \\ & = \oint\limits _ {\partial \Omega} (u dx + - v dy) + i \oint\limits _ {\partial \Omega} ( u dy + v dx) & \\ & = \iint\limits _ {\Omega} (- u _ y - v _ x ) dx\ dy + i \iint\limits _ {\Omega} ( u _ x - v _ y ) dx \ dy & \\ & = 0 + 0i = 0 & \\ & & \blacksquare \\ \end{array} $$

Singularities

Type I (removable)

$$ \text{Def: } \lim\limits _ {z \to z _ 0} f(z) = \omega _ 0 $$

example

$$ \frac{\sin(z)}{z} $$

$z=0$ undefined but:

$$ \frac{1}{z}(z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots ) = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \dots = 1 $$

Type II (pole)

$$ \text{Def: } \lim\limits _ {z \to z _ 0} |f(z)| = + \infty $$

example

$$ \frac{1}{z} $$

Type III (essential)

Def: neither of the other two

example

$e^{\frac{1}{z}}$, $\ln(z)$

Picards Theorem

(stated without proof)

Every punctured disc about an isolated essential singularity assumes every complex value with at most one exception.

Riemann Mapping Theorem

$\Omega$ domain in $\mathbb{C}$. There exists a 1-1 analytic (and conformal) mapping $f$ of $\Omega$ onto the open unit disc iff $\Omega$ is simply connected but not equal to the entire plane $\mathbb{C}$.

Misc.

$$ \begin{array}{ll} \oint\limits _ {C(0,1)} \frac{1}{z} dz & = \oint \frac{x - iy}{x^2 + y^2} (dx + idy) \\ & = \int\limits _ {0} ^ {2 \pi} (\cos(t) - i \sin(t)) (-\sin(t) + i \cos(t)) dt \\ & = \int\limits _ {0} ^ {2 \pi} (-cos(t)\sin(t) + \cos^2(t)) dt \\ & = 2 \pi i \end{array} $$

$\frac{1}{z}$ is not analytic on interior of $C(0,1)$ (which is why Cauchy's theorem doesn't apply).

$$ \begin{array}{l} \text{Res}(f(a)) = \lim\limits _ {z \to a} (z - a) f(z) \\ z = r e^{i \theta} \to \text{Arg}(z) = \theta \\ \end{array} $$

$\text{Res}$ for Residue.

$f(z)$ not analytic at $z _ 0$. Assume $g(z)$ analytic in $\Omega$.

$$ \oint\limits _ {\partial \Omega} \frac{g(z)}{z - z _ 0} dz = 2 \pi g(z _ 0) $$

So,

$$ \oint\limits _ {\partial \Omega} \frac{g(z)}{z - z _ 0} dz = 2 \pi \text{Res}(f(z _ 0)) $$

in general,

$$ \boxed{ \oint\limits _ {\partial \Omega} f(z) dz = 2 \pi i \sum\limits _ k \text{Res}(f(a _ k)) } $$

$f(z)$ continuous on $\Gamma$, $|f(z)| \le M$ on $\Gamma$ and $L = $ length $\Gamma$

$$ \to \left| \int\limits _ {\Gamma} f(z) dz \right| \le \int\limits _ {\Gamma} |f(z)||dz| \le ML $$

(ML inequality)

$f(z)$ analytic on $\Omega$, continuous on $\overline \Omega$

$$ \to \text{max}( |f(z)|) \in |f(w)|,\ w \in \partial \Omega $$

(weak maximum modulus principle)

$$ n \in \mathbb{Z} \to \oint\limits _ C z^n dz = \left\{ \begin{array}{ll} 0 & n > 0 \\ 2 \pi i & n = 0 \\ 0 & n < 0 \\ \end{array} \right. $$

$$ \begin{array}{ll} p(z) & = \sum p _ j z^j \to p(z) z ^ {-k-1} = \sum p _ j z ^ {j - k -1} \\ \to & \oint\limits _ C p(z) z^{-k-1} dz = \oint\limits _ C \sum p _ j z^{j-k-1} dz \\ & = \sum \oint\limits _ C p _ j z^{j-k-1} dz \\ & = p _ k \\ \to & p _ k = \oint\limits _ C p(z)z^{-k-1} dz \end{array} $$

2025-05-24