Coursera: Analysis of a Complex Kind. 5.1 Introduction to Complex Integration

Jul 30, 2016

 

Slide 2 Real integration

Start with Riemann integral as sum of rectangles between curve and x-axis.

Slide 3 Fundamental Theorem of Calculus

Slide 4 Properties of Integrals

Definition of antiderivative

Slide 5 Integral f(z) based on real integrals

Real integral is over an interval [a,b].

Complex integral is over a curve in C.

One part of the process is to create the curve from the (real) interval.

A (Complex) curve is a (piecewise) smooth function, γ, that maps a real interval into the complex plane

In symbols: γ [a,b] -> C, γ = x(t) + i y(t)

Notice t is a real value that lies in the interval [a,b] and x and y are functions of a real variable (t).

Next we have (are given) a function f that has domain γ.

\int_{\gamma} f(z) \; dz =  \underset{n \to \infty}{\lim} \; \overset{n-1}{\underset{j=0}{\Sigma}} \; f(z_j)(z_{j+1} - z_j)

{to remember: think about f(z_j) as being the y-ordinate or function value and (z_{j+1} - z_j)  being the \delta x}

Slide 6. Path integral

\int_{\gamma} f(z) \; dz is read as the integral of the function f(z) along the path γ.

γ is the path of complex values by applying the function γ to the real interval [a,b]

If t is the real variable that runs from a to b along the interval [a,b] then it turns out (See later)

\int_{\gamma} f(z) \; dz \; = \; \int_{t=a}^{t=b} f(\gamma(t)).\gamma\,{'}(t) \; dt

So, if we are given a function, \gamma(t) that maps the real interval [a,b] to some path of complex values in the complex plan , C, where the end points of the path are \gamma(\mathbf{a}) \textrm{ and }\gamma(\mathbf{b}) respectively, then a and b become the lower and upper limits of the transformed integral.

The integral with respect to z has become an integral with respect to t, where t has real end limit values, but the transformed function f(\gamma(t)).\gamma \; {'} (t) may include “i” e.g. 2e^{it} - \frac{2}{t}. Then integrate the transformed function w.r.t. variable t and substitute the limit values.

<< rest of slide is sketch of proof >>

Slide 7 Integrals over complex valued functions

Function g has domain a real interval [a,b] and creates complex outputs based on the functions , u v, with real domain interval [a,b]

Define:

g(t) = u(t) + iv(t)

then

\int_{a}^b g(t) \; dt \; = \; \int_{t=a}^{t=b} u(t) \;dt \; + \;i\int_{t=a}^{t=b} v(t) dt

Examples

\int_{0}^{\pi} e^{it} \; dt \; = \; \int_{0}^{\pi} cos(t) \; dt \; + \; i \int_{0}^{\pi} sin(t) \; dt

= [sin t] \underset {0}{\pi} – i [cos t] \underset {0}{\pi} = 2i

or directly

\int_{0}^{\pi} e^{it} \; dt \;  = \; [-ie^{it}] \underset {0}{\pi} = 2i

Another example (direct)

\int_{0}^{1} t+i \; dt \;  = \; [\frac{1}{2}t^2 \; + \; it] \underset {0}{1} \; = \; \frac{1}{2} \: + \; i

Slide 8 Example of path integral

\gamma(t) = t + it, 0\leq t \leq 1, \gamma \; ' (t) = 1 + i , f(z) = z^2

\int_{\gamma} f(z) \; dz \;= \; \int_{0}^{1} f((\gamma(t))\gamma\;'(t)) \; dt \;  =\; \int_{0}^{1} (t + it)^2(1 + i) \; dt \;

\int_{0}^{1} {2it^2 - 2t^2} \; dt \; = \; \frac{2}{3} (-1 \; + \;i)

Slide 9 Example of path integral – changing path to a function of t

Represent |z| = 1 as path \gamma(t) \; = \;  e^{it}, 0 \leq t \leq 2 \pi

\int_{|z| = 1} \; \frac{1}{z} \; dz

\gamma \; ' \;(t) \; = \;  ie^{it}

So \int_{|z| = 1} \; \frac{1}{z} \; dz = \int_{0}^{2 \pi}  \frac{1}{\gamma(t)}\;\gamma \; ' \;(t) \; dt \;

\int_0^{2 \pi} \frac{1}{e^{it}}\;ie^{it} \; dt \; = \; i \int_0^{2 \pi}  \; dt \; = \; 2i \pi

Slide 10 Example Path integral

We are given the path function \gamma(t) = e^{it}, and told the initial and final values of the real variable, t, 0 \leq t \leq 2 \pi. This is equivalent to the representation of the path as |z| = 1. gamma will map t = [0,2 \pi] to 1 to 1 (a closed path). Differentiate gamma to create \gamma \; ' (t) = ie^{it}

So, \int_{|z|=1}z \; dz = \int_{t=0}^{t=2 \pi}\gamma \;  (t).\gamma \; ' (t) \; dt = \int_{t=0}^{t=2 \pi}e^{it}.ie^{it} \; dt

i\int_{t=0}^{t=2 \pi}e^{2it} \; dt = \frac{1}{2}[e^{2it}]_{t=0}^{t=2 \pi}= \frac{1}{2}[e^{4i\pi}- e^{0}]=0

Slide 11 example

Slide 12 General solution

\int_{|z|=1}z^m \; dz = 2i\pi \textrm{ when m = -1, = o otherwise}

 

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