Page 56 - Calculus with Complex Numbers

Proof We have

fl g(c) + (z - c)g'(c) + . . . k(z)
z) = , z/
(z - c)h (c) + (c - c) ;,,(c)ya! .y. . . . = az - c ,
where
k( # (c) + (Z - c)#J(c) + '-
z) =z ? ?? z! o .
h (c) + (z - c)h (c)( . . .
Therefore the residue of flz) at z = c is g (c)//l/ (c) by the cover up rule.

Example Consider again
1 1 1 1 1 1
Rej 2 = = - , Res = = - .
:
:=2 C + 1 2.: i 2.ï i .,2 + 1 2 . : i V
:
=
N otes
A rigorous treatment of contour integration would present the facts in a different
order. W e have assumed in our proof of the residue theorem that a differentiable
function has a valid Laurent expansion near an isolated singularity, and that this
expansion can be integrated term by term . W e have also assumed in otlr statement
of Cauchy's theorem that the çinside' of a closed contour is well delined. A rig-
orous proof of the residue theorem requires a knowledge of uniform convergence.
A rigorous proof of Cauchy's theorem requires a knowledge of plane topology.
80th of these can be found in Knopp (1945).

Exam ples

Hvaluate the following contour integrals.
'
;
(i) J Re z dz where y is the unit circle z = eit (0 :jq t :jq 2zr) .
(ii) J I z I2 dz where y is the parabolic arc z = t + itl (0 :jg t :jg 1).
;
'

(iii) J i dz where y is the straight line joining 0 to 1 + i .
;
'
Use the estimate lemma to prove the following inequalities.