Math 43 Midterm 2

winding number

(# counterclockwise) - (# clockwise)

Proof: If all loop integrals of f in D vanish, then contour integrals of f are independent

1.) Assume every loop integral of f in D vanishes. Then say γ₁ and γ₂ are two contours in D sharing the same initial and terminal points 2.) Define Γ to be the contour γ₁ - γ₂, so Γ is a loop. Therefore, integral ∫Γf(z)dz = 0 3.) ∫Γf(z)dz = ∫γ₁f(z)dz - ∫γ₂f(z)dz = 0 4.) ∫γ₁f(z)dz = ∫γ₂f(z)dz

Theorem 10: In a simply connected domain, an analytic function

1.) has an antiderivative 2.) has contour integrals that vanish 3.) has loop integrals that vanish

For a closed contour γ not passing through 0, ∫γ(1/z)dz =

2πik where k = winding #

Definition simply connected domain

A simply connected domain is any domain possessing the property that every loop in D can be continuously deformed in D to a point (simply connected domains have no holes)

If any simple closed curve lying entirely in a domain D can be deformed to a single point in D, then

D is a simply connected domain

Cauchy's Integral Theorem

If f is an analytic function in a simply connected domain D and γ is any closed contour in D, then ∫γf(z)dz = 0

ML Inequality

If f is continuous on the contour Γ and |f(z)|≤M for all z on Γ, then we can estimate the contour integral |∫f(z)dz| ≤ Ml(Γ)

Why are winding numbers important?

If γ₀∼γ₁ in D, then γ₀ and γ₁ must have the same winding number

Cauchy's Theorem 1st version

Let D be a domain and let γ be a simple closed contour in D Let f: D→C be analytic inside and on γ Then, ∫f(z)dz = 0

definition contractible

Let D be a domain and γ be a closed contour in D Then γ is contractible if it is homotopic to a point (γ∼{pt})

Fundamental theorem of line integrals

Let f be a continuous function in a domain D. TFAE 1.) f has an antiderivative, F(z) in D 2.) All loop integrals of f in D vanish (∫f(z)dz = 0 for all loops γ in D) 3.) contour integrals of f are independent of path in D (∫γ₁f(z)dz = ∫γ₂f(z)dz)

Cauchy's Main Theorem (Deformation Invariance)

Let f be a function analytic in a domain D containing the loops γ₀ and γ₁ If γ₀∼γ₁ in D, then ∫γ₀f(z)dz = ∫γ₁f(z)dz

Cauchy Integral Formula for derivatives

Let γ be a simple closed contour positively oriented. If f is analytic in some simply connected domain containing γ and if z₀ is inside γ, then fⁿ(z₀) = (n!/2πi)∫f(z₀)/(z-z₀)ⁿ⁺¹ dz

Cauchy Integral Formula

Let γ be a simple closed contour positively oriented. If f is analytic in some simply connected domain containing γ, and if z₀ is inside γ, then f(z₀) = (1/2πi)∫f(z)/(z-z₀)dz

Definition homotopic

The loop γ₀ is said to be homotopic or continuously deformable to the loop γ₁ in the domain D if ∃ a function z(s,t) continuous on the unit square 0≤s≤1, 0≤t≤1 that satisfies 1.) z(0, t) = γ₀ 2.) z(1, t) = γ₁ 3.) z(s, 0) = z(s, 1)

When f has an antiderivative throughout a domain D,

Then, 1.) integrals along a contour in D depend only on their endpoints (integrals are path independent) 2.) All loop integrals of f in D vanish

If D is any domain γ is a simple closed contour, and if f is a function that is analytic inside and on γ, then we can define

We can define a simply connected domain containing γ

simple closed contour

a closed contour with no multiple points other than its initial-terminal point

Domain is simply connected iff (2 way ↔)

all closed contours in D are contractible

antiderivative implies

analytic! because only analytic functions are contenders to have an antiderivative

Jordan Curve Theorem

any simple closed contour separates the plane into two domains, each having the curve as its boundary. Interior domain is bounded, and exterior domain is unbounded

analyticity implies

continuity but not the other way around!

If f is continuous in D and and if ∫f(z)dz = 0 for every closed contour in D, then

f is analytic in D

If f is analytic inside and on γ, a simple closed contour, then

f must be analytic in some simply connected domain containing γ, so ∫γf(z)dz = 0

If f is analytic in a domain D, then its antiderivatives

f', f'', f''',.... exist and are analytic in D as well

logarithm is not defined

for negative numbers

in a simply connected domain, a function that is analytic

has an antiderivative

∆γarg(z) =

how much the argument changes as you travel γ. ∆γarg(z) = 2kπ where k= winding #

length of a smooth curve, l(γ)

if z(t) is parameterization of γ, then l(γ) = ∫|z'(t)|dt

homotopies in entire complex plane

in entire complex plane, any loop is homotopic to every other loop

A continuous function has an antiderivative in D iff (2 way↔)

its integral around EVERY LOOP is zero

If γ is a simple closed contour, what do we know about its interior?

its interior is a simply connected domain

Cauchy Bound

let f be analytic inside and on a circle of radius R centered about z₀. If |f(z)|≤M for all z on the circle, then |fⁿ(z₀)|≤ (n!M)/Rⁿ

when interior domain is to the right of travel

negatively oriented

When interior domain is to left of direction of travel

positively oriented

analytic does NOT imply

that a function has an antiderivative - see f(z) = 1/z

z(s, 0) = z(s, 1) tells us

that each fixed s parameterizes a loop

Liouville's Theorem

the only bounded entire functions are constant functions

If a function is continuous on a directed smooth curve γ

then this means f is integrable along γ

Corollary of Theorem 6: If f is continuous in a domain D and f has an antiderivative in D,

then ∫f(z)dz = 0 for all loops γ in D.

if γ is parameterized by z(t) a≤t≤b, then -γ is parameterized by

z(-t) -b≤t≤-a

If z(s,t) parameterizes γ₀ into γ₁, then how do we parameterize γ₁ into γ₀?

z(1-s, t) parameterizes γ₁ into γ₀

parameterization of vertical line segment from z = a+ bi to z = a-di

z(t) = a + it -d≤t≤b

parameterization of function y=f(x)

z(t) = t + if(t) 0≤t≤1

parameterization of horizontal line from z=a to z=b

z(t) = t a≤t≤b

parameterization of circle of radius r centered at z₀

z(t) = z₀ +re^it 0≤t≤2π

parameterization of straight line segment from z₁ = a+bi to z₂ = c+di

z(t) = z₁ + t(z₂-z₁) 0≤t≤1

definition smooth γ

γ is smooth if it has a parameterization z(t) and the following hold: 1.) z(t) has a continuous derivative z'(t) 2.) z'(t) never vanishes 3.) z(t) is one-to-one

Theorem 6: If complex valued function f(z) is continuous in domain D and has an antiderivative throughout D, (know proof!)

∫f(z)dz = F(b) - F(a) AND ∫f(z)dz = 0 for all loops γ in D. proof of part 1: 1.) Assume γ smooth curve with paramaterization z(t) a≤t≤b 3.) Let antiderivative of f be F, so ∫f(z)dz = ∫F'(z)dz = ∫F'(z(t))z'(t)dt = ∫d/dt (F(z(t))dt 4.) by fundamental theorem of calculus, ∫d/dt(F(z(t))dt = F(z(b)) - F(z(a))`