Page 54 - Calculus with Complex Numbers

as required.

General case lf f (z) has a singularities at z = cl , . . . , cn inside y then we can
draw circles n , . . . , yn with centres at cl , . . . , cn and with radii small enough
to ensure they all lie inside y and that they don't intersect each other. Hence by
Corollal'y 3 of Cauchy's theorem we have

J'lz) dz - X J'lz) dz - X lxiRk
/ k=1 /k k= 1

by what we have already proved.

Example Hvaluate the integral

dz
. :2 + 1 '
;'
where y is to be specilied.
Answer W e need to lind the singularities of the integrand and lind the residues
at these singularities. ln fact we already did this in Section 3.9 where we found
that the singularities are at z = Ljzi with residues ulu 1/2ï .
y == circle centre i, radius 1.
lrherefore

+

z
.
1
p
j zJ2 - l.n.i (s1 ) - n..

y == circle centre --ï, radius 1. 1n this case --ï is irside y, i is outside y.
lrherefore

1
+

p
j zJ2 - zn.i (- sl ) - -n..

z
.