Coursera Analysis Complex Kind Lesson 5.5 Consequences of Cauchy’s Theorem and Integral Formula

Aug 3, 2016

Slide 2 Cauchy Theorem and Integral formula

Relating the kth derivative and the integral over a closed curve \gamma of a function of the form \int_{\gamma}\frac {f(z)}{{(z-w)}^{k+1}}

D is a simply connected domain (no holes), F is analytic in a region that contains D, then,

f^{(k)}(w) = \frac{k!}{2 \pi i}\int_{\gamma}\frac {f(z)}{{(z-w)}^{k+1}}dz for all w in D, k > 0.

Slide 3 Cauchy’s estimate

Slide 4 Liouville’s Theorem

f is analytic in the complex plane (i.e. entire function)

and if f is bounded

then f must be constant.

converse : There are no bounded, varying analytic functions in C.

c.f. Sin x in reals where function value is bounded between -1 and +1 but is (definitely) not constant.

Proof: uses Cauchy’s estimate

Slide 5 Example of Liouvile Theorem

Slide 6 Liouville Theorem to prove fundamental theorem of algebra

Slide 7 Consequence of fundamental theorem of algebra

A polynomial can be factored into n factors.

Slide 8 Maximum Principle *Important *

Let f be analytic in domain D

and suppose there exists a point z_0 in D such that f(z) \leq f(z_0) for all z in D

i.e. there is a maximum value at some point in D,

then f is constant in D.

Consequence if f is analytic on D then |f(z)| reaches its maximum on the boundary \partial D

 Slide 9 Maximum Principle Example *Important *

<< diagram >>

Domain is square where one corner is origin other corners are 1, i and 1+i, so sides of length 1. z = x + iy, so both x and y are real values between 0 and 1.

f(z) = z^2 - 2z (which is analytic but not constant)

What is the maximum of |f(z)|?

Boundary \gamma_1 , 0 \leq x \leq 1, y = 0

so |f(z)| = |f(x)|= |x^2 – 2x| = |x(x-2)|

for the domain of x, maximum occurs when x = 1, so |f(z)| \leq |f(1)| = 1 \textrm{ on } \gamma_1

Boundary \gamma_2 , 0 \leq y \leq 1, x = 1

so |f(z)| = |f(1+iy)|= |-1-y^2| = y^2 + 1

for the domain of y zero to 1 and fixed value of x=1, maximum occurs when y = 1, so |f(z)| \leq |f(1+i)| = 2 \textrm{ on } \gamma_2

<< provide detailed working for further two boundaries >>

\gamma_3 \; y = 1, 0 \leq x \leq 1

By considering the modulus of the function applied to x + i (which is one way of describing this boundary) and using pythag to get expression for the modulus of f(x+i)

e.g. wolfram gives max((x^2 -2x -1)^2 +(2x-2)^2)^{0.5}, \textrm { for x=0 to 1, is } \sqrt{5} \textrm{ at } x = 0

\gamma_4  \; x = 0, 0 \leq y \leq 1



Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: