Coursera Complex Analysis Lesson 7.4 Finding Residues

Aug 19, 2016

IMPORTANT

Slide 2 Restate Residue Theorem

  • f has isolated singularity at z_0 and is analytic on the punctured disc 0 < | z- z_0| < r
  • f has a unique Laurent series representation …
  • The residue of f at z_0 is represented by Res(f,z_0) = a_1 the coefficient of \frac{1}{z - z_0}

Following slides are to calculate residues at removable singularities and poles (not essential singularities?)

Slide 3 Residue at Removable singularity

z_0 is a removable singularity of a Laurent series if there are no negative powers of the terms (z -z_0)^k

so a_1 = 0

so Res(f,z_0) = 0

Slide 4 Residue at simple pole

f(z) = \frac{a_{-1}}{z-z_0} + a_0 + a_1(z-z_0)^1 + a_2(z-z_0)^2 +  ... (*)

How to isolate a_1? Cross-multiply then take limit as z tends to z_0

(z-z_0)f(z) ={a_{-1}} + a_0(z-z_0)^1 + a_1(z-z_0)^2 + a_2(z-z_0)^3 + ...

Res(f, z_0) = \underset{{z \to z_0}}{\lim}(z-z_0)f(z)

We need to take limit because of the division by z - z_0 in the original Laurent series (*)

Slide 5 Residue at simple pole example

(From previous slide) Res(f, z_0) = \underset{{z \to z_0}}{\lim}(z-z_0)f(z)

Consider f(z) = \frac{1}{z^2 + 1} has simple poles at z_0= +i, -i

Res(\frac{1}{1+z^2}, i) = \underset{{z \to i}}{\lim}(z-i)\frac{1}{z^2 + 1}

 = \underset{{z \to i}}{\lim}(z-i)\frac{1}{(z+i) (z-i)} = \underset{{z \to i}}{\lim}\frac{1}{(z+i)} = \frac{1}{2i}

Slide 6 Residue at double pole example

Laurent series: f(z) = ...+ \frac{a_{-2}}{(z-z_0)^2} + \frac{a_{-1}}{z-z_0} + a_0 + a_1(z-z_0)^1 + a_2(z-z_0)^2 +  ... (*)

z_0 is double pole when a_{-2} \neq 0 \textrm{ and } a_{-k} = 0 for k > 2.

To isolate a_{-1}. (i) Multiply by (z –  z_0)^2 and (ii) differentiate (iii) take limit.

f(z) = \frac{a_{-2}}{(z-z_0)^2} + \frac{a_{-1}}{z-z_0} + a_0 + a_1(z-z_0)^1 + a_2(z-z_0)^2 +  ...

(z-z_0)^2f(z) = {a_{-2}} + {a_{-1}}{(z-z_0)}^1 + a_0(z-z_0)^2 + a_1(z-z_0)^3 + a_2(z-z_0)^4 +  ...

\frac{d}{dz}(z-z_0)^2f(z) =  {a_{-1}} + 2a_0(z-z_0)+ 3a_1(z-z_0)^2 +   ...

Res(f, z_0) = a_{-1}= \underset{{z \to z_0}}{\lim}\frac{d}{dz} (z-z_0)^2f(z)

Slide 7 Residue at double pole example

f(z) = \frac{1}{(z-1)^2(z-3)} double pole at z = 1, simple pole at z = 3.

Res(\frac{1}{(z-1)^2(z-3)}, 1) =  \underset{{z \to 1}}{\lim}\frac{d}{dz} (z-1)^2\frac{1}{(z-1)^2(z-3)}

=  \underset{{z \to 1}}{\lim}\frac{d}{dz} \frac{1}{(z-3)}

=  \underset{{z \to 1}}{\lim} \frac{-1}{(z-3)^2} =-\frac{1}{4}

Slide 8 Residue at poles of order n

Use method above: (i) Multiply by (z –  z_0)^n and (ii) differentiate multiple times (iii) take limit.

Giving

Res(f,z_0) = \frac{1}{(n-1)!}\underset{{z \to z_0}}{\lim}\frac{d^{n-1}}{dz^{n-1}}[(z -  z_0)^n f(z)]

Slide 9 Residue of “quotient” functions

f(z) = \frac{g(z)}{h(z)}

Res(f,z_0) = \frac{g(z_0)}{h\; '(z_0)}

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