Coursera Complex Analysis Lesson 7.3 Residue Theorem

Aug 16, 2016

God this stuff goes on and on

Slide 2

Recap: f has an isolated singularity at $z_0$ e.g. when z = $z_0$, we have division by zero somewhere.

Then if function is analytic everywhere in a (Punctured) disc around $z_0$ (but excluding only the point $z_0$) since a punctured disc is an annulus, the function has a Laurent series representation:

f(z) = $\Sigma_{k=-\infty}^{+\infty}a_k(z - z_0)^k, \; 0 < |z - z_0| < r$  (inequality excludes $z_0$ the centre of disc).

Now take some closed circular path somewhere in the punctured disc, 0 < $\rho$ < r.

Then integrate the function around the disc, $\int_{|z - z_0|=\rho}f(z) \;dz =\Sigma_{k=-\infty}^{+\infty}a_k \int_{|z - z_0|=\rho}(z - z_0)^k \;dz\;, \; 0 < |z - z_0| < r$ (*) (not explained, that because convergence is uniform over disc then sigma and integral can be interchanged). Note we have taken $a_k$ outside the integral as it is a constant coefficient for each integral. And in theory we have to perform an infinite number of integrations as k ranges between – and + infinity.

How do we calculate these integrals, $\int_{|z - z_0|=\rho}(z - z_0)^k \;dz$ (which are contained within the sigma)?

Slide 3

For all values of k EXCEPT -1, the function $h(z) = (z - z_0)^k \textrm{ has a primitive, } H(z) = \frac{1}{(k+1)} (z - z_0)^{k+1}$

and since we are integrating an (analytic?) function over a closed curve then $\int_{|z - z_0|=\rho}(z - z_0)^k \;dz\;=\;0$, see,

So almost all the terms that are integrated have disappeared.

For the remaining integral $\int_{|z - z_0|=\rho}(z - z_0)^{-1} \;dz\; = \int_{|z - z_0|=\rho}\frac{1}{(z - z_0)} \;dz\;$ we can parameterise the curve , or use the Cauchy Integral theorem, to give a value of $2i\pi$

And because this integral value is for k=-1 (all other integral being zero), looking back at (*) the final value for the sigma is $a_{-1}2i\pi$

Slide 4

From (*) above, at an isolated singularity, $z_0$, the Residue is the $a_{-1}$ coefficient of the Laurent expansion of f(z) around the isolated singularity $z_0$, all other coefficients being zero.

If annulus around isolated singularity $z_0 \textrm{ is } 0<|z-z_0| the the residue of f at $z_0$ is represented as $Res(f,z_0) = a_{-1}$ (of the Laurent expansion of f around z_0).

examples

f(z) = $\frac{1}{(z-1)(z-2)}$

This has isolated singularities at z = 1,2

Therefore annuli around the singularities

0 < |z-1| < 1 and 0 < |z-2| < 1

Laurent series

f(z) = $-1(z-1)^{-1}$+ sum rest of terms

Res(f,1) = -1

or

f(z) = $1(z-2)^{-1}$+ sum rest of terms

Res(f,2) = +1

Slide 5 Residue examples $f(z) = \frac{sin z}{z^4} = \frac{1}{z^3} -\frac{1}{3!} \; \frac{1}{z} + \frac{1}{5!} z - ...$

Isolated singularity z = 0

Laurent series over annulus 0 < | z – 0 | < $\infty$

So Res(f,0) = $- \frac{1}{3!}$

• f(z) = cos(1/z) = ….

this is an essential singularity (see) because infinitely many terms with negative powers $cos (\frac{1}{z}) = 1 - \frac{1}{z^2}\frac{1}{2!} + \frac{1}{z^4}\frac{1}{4!} - \frac{1}{z^6}\frac{1}{6!} + ...$

so z = 0 is an essential singularity and since coefficient of z^-1 is zero

Res(f,0) = 0

• f(z) = sin(1/z) = ….

this has an essential singularity (see) because infinitely many terms with negative powers $sin (\frac{1}{z}) = \frac{1}{z}\frac{1}{1!} - \frac{1}{z^3}\frac{1}{3!} + \frac{1}{z^5}\frac{1}{5!} + ...$

so z = 0 is an essential singularity and since coefficient of z^-1 is 1

Res(f,0) = 1

• f(z) = $\frac{-1 + cos z}{z^2} = ....$

this has a removable singularity (see) because no terms with negative powers $\frac{-1 + cos z}{z^2} = -\frac{1}{2!} + \frac{z^2}{4!} - ...$

so z = 0 is an removable singularity and since coefficient of z^-1 is 0

Res(f,0) = 0

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