236 CHAPTER 6 • COMPLEX INTEGRATION
Figure 6.33 The contours G and Co in the proof of Cauchy's integral formula.• EXAMPLE 6. 21 Show that fct(o) ";~{ dz = i27re.
Solution Recall tha t ct (0) is the circle centered at 0 with radius 1 and having
positive orientation. We have f (z) = exp z and f ( 1) = e. The point zo = 1 lies
interior to the circle, so Cauchy's integral formula implies that
e =/(1) = -^1 1 expz
2
. --
1
dz,
7rt c z -and multiplication by 2m establishes the desired result.
- EXAMPLE 6.22 Show that fct(o) :!~: dz= i-f" ·
Solution Here we have f(z) = sinz. We manipulate the integral and use
Cauchy's integral formula to obtain
r sinz dz= ~ r sinz dz
lct 4z + 1T 4 l ct z + (~)
= ~ { I (z) dz
4 lct<o> z - (-.n= ~ (27ri) I ( ~1r)
= ~i sin ( -47r) = -v:1Ti.
•EXAMPLE 6.23 Show that fct(o} 2 :;~;:1 2 dz=^23 ".Solutio n We see that 2z^2 - 5z + 2 = (2z - 1) (z - 2) = 2 (z - ~) (z -2). The
only zero of this expression that lies in the interior of C1 (0) is zo = ~. We set