Complex Analysis

complex number $C$

$z = (x, y)$ in complex plane where $x, y \in R$

real part, $x = Re z$
imaginary part, $y = Im z$

complex plane

real axis/ imaginary axis
$(0, y)$ pure imaginary number, $y \neq = 0$

binary operation of complex number

addition, $(x_{1}, y_{1}) + (x_{2}, y_{2}) = (x_{1} + x_{2}, y_{1} + y_{2})$
multiplication, $(x_{1}, y_{1}) \cdot (x_{2}, y_{2}) = (x_{1} x_{2} - y_{1} y_{2}, x_{1} y_{2} + x_{2} y_{1})$

complex number short syntax

$(x, y) = (x, 0) + (0, y) = (x, 0) + (0, 1) \cdot (y, 0) =: x + i y$

where $i := (0, 1)$

Argand diagram

diagram in $R^{2}$ given by

$x + i y \mapsto x i + y j$

modulus, $=$ Euclidean length in Argand diagram

$∣ z ∣ = x^{2} + y^{2}$
triangle inequality hold

complex conjugation

$z ˉ = x - i y$

$z ˉ ˉ = z$
$∣ z ˉ ∣ = ∣ z ∣$
$z_{1} + z_{2} = z ˉ_{1} + z ˉ_{2}$
$z_{1} z_{2} = z ˉ_{1} z ˉ_{2}$
$x = \frac{z + z ˉ}{2}, y = \frac{z - z ˉ}{2 i}$
$∣ z ∣^{2} = z z ˉ$

Euler’s formula

$e^{i θ} = cos θ + i sin θ \Rightarrow x + i y = r e^{i θ}$

$r = ∣ z ∣$
$θ = ar g z$ , argument of $z$
- $- π < Arg z \leq π$ , principal argument of $z$
$z_{1} z_{2} = r_{1} r_{2} e^{i (θ_{1} + θ_{2})}$
- $ar g (z_{1} z_{2}) = ar g z_{1} + ar g z_{2}$
$z^{- 1} = \frac{1}{r} e^{- i θ}$
$z ˉ = r e^{- i θ}$

de Moivre’s formula

$(cos θ + i sin θ)^{n} = e^{in θ} = cos n θ + i sin n θ$

complex root

$z_{0} = r_{0} e^{i θ_{0}}$

$n$ th root of $z_{0}$

$z^{n} = z_{0} \Rightarrow z_{k} = n r_{0} e^{i (\frac{θ _{0} + 2 π k}{n})}, k \in {0, 1, \dots, n - 1}$

$n$ distinct root
principal root, when $θ_{0} = Arg z_{0}, k = 0$

standard topology $O_{s}$

see context in Topology

$ε > 0$

open $ε$ -neighborhood

$U_{ε} (z_{0}) = {z \in C : ∣ z - z_{0} ∣ < ε} \in O_{s}$

region

nonempty connected set

domain, open region

connectedness

$D \subseteq C$ is connected $\Leftrightarrow$

$\forall z_{1}, z_{2} \in D, \exists$ polygonal line with finitely many segment between $z_{1}$ , $z_{2}$

bounded

$\exists$ circle of finite radius containing the set

Riemann sphere $S$

$x^{2} + y^{2} + z^{2} = 1$

intersection of the sphere with line through north poll and complex number

north poll: $(0, 0, 1)$ , correspond to complex infinity
Argand plane $(x, y, 0)$

limit

$z \to z_{0} lim f (z) = w_{0}$

$\Leftrightarrow \forall ε > 0, \exists δ > 0$ s.t.

$∣ f (z) - w_{0} ∣ < ε whenever ∣ z - z_{0} ∣ < δ$

unique if exist
independent of direction of approach

connection between complex limit and real limit

$f (z) = u (x, y) + i v (x, y)$

$z \to z_{0} lim f (z) = (x, y) \to (x_{0}, y_{0}) lim u (x, y) + i (x, y) \to (x_{0}, y_{0}) lim v (x, y)$

$\Leftrightarrow$ limit on the RHS exist

continuity

$f$ is continuous at $z_{0} \Leftrightarrow$

$f$ exist
$lim_{z \to z_{0}} f (z)$ exist
$z \to z_{0} lim f (z) = f (z_{0})$

differentiation

$f : D \to C, U_{ε} (z_{0}) \subseteq D$

$f$ differentiable at $z_{0} \Leftrightarrow$

$z \to z_{0} lim \frac{f ( z ) - f ( z _{0} )}{z - z _{0}}$

exist

$f (z)$ depend on $z ˉ \Rightarrow f$ not differentiable
$u, v$ differentiable $\neq \Rightarrow f$ differentiable

Cauchy-Riemann equations

$f^{'} (z)$ exist at $z_{0} \Rightarrow$

$u_{x} = v_{y}, u_{y} = - v_{x} f^{'} = u_{x} + i v_{x}$

Cauchy-Riemann equations in polar coordinate

$f (r, θ) = u (r, θ) + i v (r, θ)$

$r u_{r} = v_{θ}, u_{θ} = - r v_{r} f^{'} = e^{- i θ} (u_{r} + i v_{r})$

differentiability from Cauchy-Riemann equations

$f$ defined around $z_{0}$ ,
at $z_{0}$

$u_{x}, u_{y}, v_{x}, v_{y}$ are continuous
satisfy Cauchy-Riemann equations

$\Rightarrow f^{'} = u_{x} + i v_{x}$

exist

holomorphicity

$f : D \to C$ holomorphic at $z_{0} \Leftrightarrow$

$f$ differentiable at each point in $U_{ε} (z_{0}) \subseteq D$

a.k.a. analytic/ regular
$u, v \in C^{\infty} (D)$ and are harmonic function in $D$

constancy from holomorphicity

$f$ holomorphic, in $D$

$f$ is constant

$\Leftrightarrow f^{'} = 0$

$\Leftrightarrow f ˉ$ holomorphic in $D$

$\Leftrightarrow ∣ f ∣$ is constant

parametric curve

complex-valued function

$C : [a, b] \subseteq R \to C, t \mapsto z (t) = x (t) + i y (t)$

simple $\Leftrightarrow$ does not intersect self $\Leftrightarrow$ injection
- simple closed (Jordan curve) $\Leftrightarrow$ simple except two end meet
oriented
- positively oriented $\Leftrightarrow$ simple closed counterclockwise
- negatively oriented

parametric derivative

$x^{'} (t), y^{'} (t)$ continuous

$\Rightarrow C$ differentiable,

$z^{'} (t) = x^{'} (t) + i y^{'} (t)$

velocity
- speed $∣ z^{'} ∣$
smooth $\Leftrightarrow$ differentiable and $z^{'} \neq = 0$
- $C$ has unit tangent vector in Argand diagram
  
  $T (t) = \frac{z ^{'} ( t )}{∣ z ^{'} ( t ) ∣}$

parametric integral

$\int z (t) = \int x (t) + i \int y (t)$

both real part and imaginary part commute
antiderivative
mean value theorem of integral fail

reparametrization

$t ~ \mapsto ϕ (t ~) = t C : z (t) = z ~ (t ~)$

curve length persist

parametric curve length

$L (C) = \int_{a}^{b} x^{'} (t)^{2} + y^{'} (t)^{2} d t = \int_{a}^{b} ∣ z^{'} (t) ∣ d t$

contour

piecewise smooth curve $C$ joined from finite smooth parametric curve

$z$ continuous
$z^{'}$ piecewise continuous
vertex, point where $C$ not smooth
length, orientation, simple closed like parametric curve

Jordan curve theorem

simple closed contour separate $C$ into

a bounded interior
an unbounded exterior

contour integral

contour $C : z (t) : [a, b] \to C$ ,
function $f (z)$ piecewise continuous on $C$

$\int_{C} f (z) d z = \int_{a}^{b} f (z (t)) z^{'} (t) d t$

independent of parametrization
linearity

$\int_{C} (c f (z) + g (z)) d z = c \int_{C} f (z) d z + \int_{C} g (z) d z$

for contour $C = C_{1} \cup C_{2}$

$\int_{C} f (z) d z = \int_{C_{1}} f (z) d z + \int_{C_{2}} f (z) d z$
traversed in opposite direction

$- C : t \mapsto z (- t) : [- b, - a] \to C \Rightarrow \int_{- C} f (z) d z = - \int_{C} f (z) d z$

upper bound theorem for contour integral (ML theorem)

contour $C$ of length $L$ ,
$f (z)$ piecewise continuous on $C$ ,
$f (z)$ bounded by $M$

$\Rightarrow \int_{C} f (z) d z \leq M L$

proof using lemma

$\int_{a}^{b} z (t) d t \leq \int_{a}^{b} ∣ z (t) ∣ d t$

proof of lemma

$I := \int_{a}^{b} w (t) d t = r_{I} e^{i θ_{I}}$

lemma hold for $I = 0$ . for $I \neq = 0$

$r_{I} > 0 \Rightarrow r_{I} = Re (e^{- i θ_{I}} \int_{a}^{b} w (t) d t) = \int_{a}^{b} Re (e^{- i θ_{I}} w (t)) d t \leq \int_{a}^{b} e^{- i θ_{I}} w (t) d t = \int_{a}^{b} ∣ z (t) ∣ d t$

path independence of contour integral

$f : D \subseteq C \to C$ continuous in $D$

$\exists$ antiderivative $F$ s.t. $\forall C$ in $D$ from $z_{1}$ to $z_{2}$

$\int_{C} f (z) d z = F (z_{2}) - F (z_{1})$

$\Leftrightarrow \forall C$ closed,

$\oint_{C} f (z) d z = 0$

Cauchy-Goursat theorem

on $D \subseteq C$ ,
simple closed contour $C$ in $D$ ,
$f : D \to C$ holomorphic on and within $C$

$\Rightarrow \int_{C} f (z) d z = 0$

Cauchy’s theorem

same as Cauchy-Goursat theorem, except $f^{'} (z)$ continuous

proof using Green's theorem and Cauchy-Riemann equations

$\int_{C} f (z) d z = \int_{C} (u + i v) (d x + i d y) = \int_{C} (u + i v) d x + (- v + i u) d y = \iint_{D} (\frac{\partial}{\partial x} (- v + i u) - \frac{\partial}{\partial y} (u + i v)) d A = \iint_{D} (- v_{x} + i u_{x} - u_{y} + i v_{y}) d A = 0$

Cauchy-Goursat theorem on simply connected domain

on simply connected domain $D \subseteq C$ ,
closed contour $C$ in $D$ ,
$f : D \to C$ holomorphic on $D$

$\Rightarrow \int_{C} f (z) d z = 0$

simply connected domain: every simple closed contour enclose only point in $D$
- multiply connected domain
holomorphic function on simply connected domain has antiderivative
- entire function has antiderivative

adoption of Cauchy-Goursat theorem on multiply connected domain

on multiply connected domain $D \subseteq C$ ,
positively oriented contour $C$ , negatively oriented simple closed contour $C_{i}$ ,
$C_{i}$ inside of and disjoint from $C$ ,
$f : D \to C$ holomorphic on $D$ and on the region between $C, C_{i}$

$\Rightarrow \int_{C} f (z) d z + \int_{C_{i}} f (z) d z = 0$

path deformation principle

for the following case, deformation of contour integral persist its value

positively oriented contour $C_{1}, C_{2}$ ,
$f$ holomorphic between $C_{1}, C_{2}$

$\Rightarrow \int_{C_{1}} f (z) d z = \int_{C_{2}} f (z) d z$

Cauchy’s integral formula (Cauchy’s formula)

$C$ positively oriented contour,
$f$ holomorphic inside and on $C$

$\forall z_{0}$ inside $C$

$f (z_{0}) = \frac{1}{2 π i} \int_{C} \frac{f ( z ) d z}{z - z _{0}}$

derivative

$f^{'} (z_{0}) = \frac{1}{2 π i} \int_{C} \frac{f ( z ) d z}{( z - z _{0} ) ^{2}}$
at $z_{0}$ , $f$ holomorphic $\Rightarrow \forall n \in N, f^{(n)}$ exist and holomorphic

$f^{(n)} (z_{0}) = \frac{n !}{2 π i} \int_{C} \frac{f ( z ) d z}{( z - z _{0} ) ^{(n + 1)}}$
$f$ holomorphic $\Rightarrow u, v$ have continuous partial derivative at all order

proof

show using upper bound theorem

$\int_{C_{R}} \frac{f ( z ) - f ( z _{0} )}{z - z _{0}} d z = 0$

evaluating integral with Cauchy’s integral formula

on positively oriented contour $C$ , evaluate integral

$I = \int_{C} g (z) d z$

find $f (z), z_{0}$ s.t. $z_{0}$ inside $C$ and

$g (z) = \frac{f ( z )}{( z - z _{0} ) ^{n}}$
calculate $f^{(n - 1)} (z_{0})$
apply Cauchy’s integral formula

$I = \frac{2 π i}{( n - 1 )!} f^{(n - 1)} (z_{0})$

Morera’s theorem

$f$ continuous on $D$

$\forall$ closed contour $C$

$\int_{C} f (z) d z = 0$

$\Rightarrow f$ holomorphic throughout $D$

Liouville’s theorem

bounded entire function is constant

fundamental theorem of algebra

non-constant polynomial of degree $n$ has $n$ root

maximum modulus principle

non-constant holomorphic function $f$ on open $D$
$\Rightarrow ∣ f ∣$ has no maximum on $D$

for $f$ on $D \cup \partial D$ , $∣ f ∣$ reach maximum always on $\partial D$ , never on interior

analyticity

$f : U_{ε} (z_{0}) \to C$ analytic $\Leftrightarrow$

$\forall z \in U_{ε} (z_{0}), f (z) = n = 0 \sum \infty a_{n} (z - z_{0})^{n}$

Laurent series

annual domain $D$ , $R_{0} < ∣ z - z_{0} ∣ < R_{1}$ ,
any closed contour $C$ in $D$ ,
function $f$ holomorphic in $D \Rightarrow$

$f (z) = n = - \infty \sum \infty c_{n} (z - z_{0})^{n} c_{n} = \frac{1}{2 π i} \int_{C} \frac{f ( z ) d z}{( z - z _{0} ) ^{n + 1}}$

punctured disk, $R_{0} = 0$
unique

residue

$r es f (z_{0}) = c_{- 1} = \frac{1}{2 π i} \int_{C} f (z) d z \Rightarrow \int_{C} f (z) d z = 2 π i r es f (z_{0})$

Cauchy residue theorem

positively oriented contour $C$ ,
$f$ analytic on and within $C$ except on finite number of singularities $z_{k}, k \in {1, 2, \dots, n}$

$\int_{C} f (z) d z = 2 π i k = 1 \sum n r es f (z_{k})$

residue at infinity

all singularity are inside negatively oriented contour $C$

$\int_{C} f (z) d z = 2 π i r es f (\infty)$

Cauchy residue theorem can be rewritten as

$k = 1 \sum n r es f (z_{k}) + r es f (\infty) = 0$
by replacing $z$ with $\frac{1}{z}$

$r es (\frac{1}{z ^{2}} f (\frac{1}{z})) (0) = - r es f (\infty)$

singularity

$z_{0}$ is singularity $\Leftrightarrow$

$f$ holomorphic in $U^_{ε} (z_{0})$ and not at $z_{0}$

isolated singularity

$f$ analytic in deleted neighborhood $U^(z_{0})$

$z_{0}$ is isolated singularity

$f$ has Laurent series about $z_{0}$
not branch point
isolated singular at infinity $\Leftarrow \exists R > 0$ s.t. $f$ analytic for $R < ∣ z ∣ < \infty$

essential isolated singularity

$\neq \exists m \in N^{+}$ s.t. $\forall n < - m, c_{n} = 0$

$z_{0}$ is essential isolated singularity

Casorati-Weierstrass theorem

$\forall ε, R > 0, w \in C, \exists z \in ∣ z - z_{0} ∣ < R$ s.t $∣ f (z) - w ∣ < ε$

pole

$\exists m \in N^{+}$ s.t. $\forall n < - m, c_{n} = 0$

pole of order $m$ at $z_{0}$

simple pole $\Leftarrow m = 1$
removable singularity $\Leftarrow m = 0$
- removed by setting $f (z_{0}) = c_{0}$

residue at pole

$r es f (z_{0}) = {\frac{1}{( m - 1 )!} \frac{d ^{m - 1}}{d z ^{m - 1}} [(z - z_{0})^{m} f (z)]} (z_{0})$

only usable for $m = 1, 2$
$m = 1$

$r es f (z_{0}) = [(z - z_{0}) f (z)] (z_{0})$
$m = 2$

$r es f (z_{0}) = {\frac{d}{d z} [(z - z_{0})^{2} f (z)]} (z_{0})$

zero of order $m$

$f$ analytic at $z_{0}$ ,
$f (z_{0}) = 0$ ,
$\exists m \in N^{+}$ s.t.

$\forall n < m, f^{(n)} (z_{0}) = 0$

identically zero $\Leftarrow m = \infty$
- otherwise, the zero is isolated

zero and pole

$p, q$ analytic,
$q$ has zero of order $m$ at at $z_{0}$ ,
$p (z_{0}) \neq = 0$

$\Rightarrow \frac{p}{q}$ has pole of order $m$ at $z_{0}$

$m = 1 \Rightarrow \frac{p}{q}$ has simple pole at $z_{0}$ ,

$r es [\frac{p}{q}] (z_{0}) = \frac{p ( z _{0} )}{q ^{'} ( z _{0} )}$

improper integral

Cauchy principal value $C P V$

$C P V \int_{- \infty}^{\infty} f (x) d x = R \to \infty lim \int_{- R}^{R} f (x) d x$

if RHS exist

$f$ is even $\Rightarrow \int_{- \infty}^{\infty} f (x) d x = C P V \int_{- \infty}^{\infty} f (x) d x$

improper integral for even rational function

real, continuous, even, irreducible rational function $f = \frac{p}{q}$ ,
$q (x)$ has finitely many zero $z_{k}$ above the real axis

$\Rightarrow f$ has pole $z_{k}$ above the real axis,

let $C_{R}$ be upper semicircle with missing side $[- R, R]$ , then

$\int_{- R}^{R} f (x) d x + \int_{C_{R}} f (x) d x = 2 π i k \sum r es f (z_{k})$

if $lim_{R \to \infty} \int_{C_{R}} f (x) d x = 0$ , then

$\int_{- \infty}^{\infty} f (x) d x = 2 π i k \sum r es f (z_{k})$

Fourier integral

$k > 0$

$\int_{- \infty}^{\infty} f (x) sin (k x) d x or \int_{- \infty}^{\infty} f (x) cos (k x) d x$

integrate instead

$\int_{- R}^{R} f (x) e^{ik x} d x$

Jordan’s lemma

$f (z)$ analytic above the imaginary axis outside $∣ z ∣ < R_{0}$ ,
semicircle contour $C_{R} : ∣ z ∣ = R > R_{0}, 0 \leq θ \leq π$

$∣ f (z) ∣ (C_{R}) \leq M_{R}, R \to \infty lim M_{R} = 0 \Rightarrow \forall k > 0, R \to \infty lim \int_{C_{R}} f (z) e^{ik z} d z = 0$

proof by parametric integral and Jordan’s inequality

Jordan’s inequality

$\int_{0}^{π} e^{- k R s i n θ} d θ < \frac{π}{k R}$

indented path

improper integral with singularity on real axis

circumvent each singularity $x_{i}$ on real axis by upper semicircle $C_{i} : ∣ z - x_{i} ∣ = r_{i}$ with $r_{i} \to 0$

lemma for indented path

$f$ analytic on $0 < ∣ z - x_{i} ∣ < r$ ,
clockwise upper semicircle $C_{i} : z = x_{i} + r_{i} e^{i θ}, π \geq θ \geq 0, r_{i} < r$ ,
Laurent series of $f$ about $x_{i}$ contain no even negative power

$\Rightarrow r_{i} \to 0 lim \int_{C_{i}} f (z) d z = - iπ r es f (x_{i})$

proof by integrating Laurent series of parametric line integral term by term

lemma hold at simple pole
can prove

$\int_{- \infty}^{\infty} \frac{sin x}{x} d x = π$

lemma for integral over branch point

circular segment of $C_{r} : z = r e^{i θ}, 0 \leq θ \leq θ_{1}$ ,
$f (z)$ continuous on $C_{r} \Rightarrow$

$z \to 0 lim z f (z) = 0 \Rightarrow r \to 0 lim \int_{C_{r}} f (z) d z = 0$

proof by definition of limit and upper bound theorem

trigonometric integral

$\int_{θ_{0}}^{θ_{1}} f (sin θ, cos θ) d θ$

define $C : z = e^{i θ}, θ_{0} \leq θ \leq θ_{1}$

$sin θ = \frac{z - z ^{- 1}}{2 i}, cos θ = \frac{z + z ^{- 1}}{2}, d θ = \frac{d z}{i z} \Rightarrow \int_{θ_{0}}^{θ_{1}} f (sin θ, cos θ) d θ = \int_{C} f (\frac{z - z ^{- 1}}{2 i}, \frac{z + z ^{- 1}}{2}) \frac{d z}{i z}$

Laplace transform

Laplace transform of $f : R_{0}^{+} \to R$

$f ~ (s) = \int_{0}^{\infty} f (t) e^{- s t} d t : C \to C$

Bromwich formula

inverse Laplace transform

$γ \in R$ to the right of all sinularity of $f ~ (s)$ ,
line contour $Γ_{R}$ from $γ - i R$ to $γ + i R$

$f (t) = \frac{1}{2 π i} R \to \infty lim \int_{Γ_{R}} f ~ (s) e^{s t} d s$

meromorphic

$f$ is analytic except possibly for poles

theorem for meromorphic function

simple closed contour $C$ ,
$f$ non-zero and analytic on $C$ , meromorphic within $C$

$\frac{1}{2 π i} \int_{C} \frac{f ^{'} ( z )}{f ( z )} d z = Z - P$

$Z$ , number of zero of $f$ within $C$ counted with order
$P$ , number of pole of $f$ within $C$ counted with order

winding number

closed contour $C : z = z (t), a \leq t \leq b, z (a) = z (b) = z_{0}$ ,
$f (z) = ρ e^{i ϕ} \neq = 0$ on $C$ ,
$Γ : f (z (t))$

difference in argument

$Δ_{C} ar g f = ϕ (b) - ϕ (a) = 2 π k$

winding number of $Γ$ with respect to $0$

$ν (Γ, 0) = \frac{1}{2 π} Δ_{C} ar g f = k$

argument principle

$Z - P = \frac{1}{2 π i} \int_{C} \frac{f ^{'} ( z )}{f ( z )} d z = \frac{1}{2 π} Δ_{C} ar g f = ν (Γ, 0)$

Rouché’s theorem (dog on a leash theorem)

$f, g$ analytic on and within $C$ ,
$∣ g ∣ < ∣ f ∣$ on $C$

$\Rightarrow f, f + g$ have the same number of zero counted with order inside $C$

constrain location of polynomial root using Rouché’s theorem
1. split polynomial $P = f + g$ where $f (z) = z^{n}$
2. $C : ∣ z ∣ = 1 + ∣ a_{n - 1} ∣ + \dots + ∣ a_{0} ∣$
  $\Rightarrow ∣ g ∣ < ∣ f ∣$ on $C$
3. $\Rightarrow$ root of $P$ are within $C$

Brouwer’s fixed point theorem special version

$D : ∣ z ∣ \leq 1, in t D : ∣ z ∣ < 1$ ,
$g : D \to in t D$ analytic

$\Rightarrow \exists! z_{0} \in D$ s.t. $g (z_{0}) = z_{0}$ (fixed point)

when changing $w$ continuously without crossing $Γ$ , $ν (Γ, w)$ is topological invariant

Hopf’s theorem

closed curve $Γ, Γ ~$ ,
one can be continuously deformed into another without crossing point $w$

$\Leftrightarrow ν (Γ, w) = ν (Γ ~, w)$

vector field index theorem

vector field $V : z \mapsto f (z)$ ,
simple closed contour $C$ enclose singularity $z_{i}, i = 1, 2, \dots$ of $V$ with index $I_{V} (C_{i}) \Rightarrow$

$I_{V} (C_{i}) = ν (Γ, 0) = Z - P = i \sum I_{V} (C_{i})$

vector field singularity

zero or pole of $f$

vector field index

positively oriented contour $C_{i}$ enclose singularity $z_{i}$ of $V$

vector field index is integer number of rotation of $V$ traversing $C_{i}$

$I_{V} (C_{i}) = ν (Γ_{i}, 0)$

Möbius transformation

$T (z) = \frac{a z + b}{cz + d} : C ˉ \to C ˉ, a d - b c \neq = 0$

extended complex plane $C ˉ := C \cup {\infty}$
only singularity is simple pole $- \frac{d}{c}$
inverse

$T^{- 1} (w) = \frac{b - d w}{c w - a}$
derivative non-zero

$T^{'} (z) = \frac{a d - b c}{( cz + d ) ^{2}} \neq = 0$
- analytic except at pole
- non-constant mapping
is composition of linear transformation and inversion

$f_{1} (z) = cz + d, f_{2} (z) = \frac{1}{z}, f_{3} (z) = \frac{a}{c} - \frac{a d - b c}{c} z \Rightarrow T = f_{1} \circ f_{2} \circ f_{3}$
they are a group
map circle to circle in $C ˉ$
- line are circle of infinite radius through $\infty$
need $3$ point to specify

circle in Argand diagram

$A z z ˉ + (B - i C) z + (B + i C) z ˉ + D = 0 A, B, C, D \in R$

cross-ratio

map $3$ distinct point $z_{1}, z_{2}, z_{3}$ to $w = T (z)$ by
$z_{1} \mapsto w_{1}, z_{2} \mapsto w_{2}, z_{3} \mapsto w_{3}$

$\frac{( w - w _{1} ) ( w _{2} - w _{3} )}{( w - w _{3} ) ( w _{2} - w _{1} )} = \frac{( z - z _{1} ) ( z _{2} - z _{3} )}{( z - z _{3} ) ( z _{2} - z _{1} )}$