6.3 • THE CAUCHY- GOURSAT THEOREM 227y y(a) The figure eight contour C. (b) The contours C 1 and Cz.Figure 6.28 The contour C = Ci + C2.
Solution Again, we use partial fractions to express the integral:
{ z-
2
dz= 2 {~dz- { -
1- dz.
le z^2 - z le z l e z - 1
(6-39)Using the Cauchy- Goursat theorem, Property (6-17), and Corollary 6.1 (with
zo = 0), we compute the value of the first integral on the right side of Equation
(6-39):2 r ~dz=2 r ~dz+2 r ~dz
l e z le, z le, z
= -21 ~ dz+ O
-c, z= -2 (211'i) = -471'i.
Similarly, we find that
fdz fdz {dz
- le z -1 = -le, z - 1 - le. z -I
= 0 - 271'i = -271'i.
If we substitute the results of the last two equations into Equation (6-39), we get
1
z-2
--dz = -471'i - 211'i = -611'i.
c z2 - z-------~EXERCISES FOR SECTION 6.3- Determine the domain of analyticity for the following functions and evaluate
fct(o) f (z) dz.
(a) f (z) = ,i':w
(b) f (z) = ,2+~•+ 2.
(c) f (z) =tan z.