· Integration 471
so that F 1 (z) = Logz + K for z E G - L 0 , K being a constant. But
F1 (1) = 0 and Log 1 = 0, hence J{ = 0, and we obtainFi(z) =Log z
Therefore, F(z) = Log z + 2k7ri = log z.If we let w = Log z and denote by g the inverse of the mapping defined
by F, we see thatz = g(w + 2k7ri)
Thus g (the inverse of the function defined by the integral 7.16-4) is a
periodic function with period 27ri.Exercises 7.2
- Evaluate each of the following integrals.
(a) 1i \z + 1)^4 dz
(b) 1n:i ez dz( c) f ii co sh 2z dz
J
e2z
( d) zZ +
1dz, C: z - 1 = eit, 0 ::; t ::; 27r
c( e) J ~ ' where c+ is any simple closed contour enclosing both
z -1
c+z = 1 and z = -1.
- Let C be a closed contour not passing through the points a, b (a f:. b ),
and suppose that !1c(a) = !1a(b). Prove that
Jez -a)-^1 (z - br^1 dz= 0
c- Let R( z) be a rational function and suppose that the denominator of
R has simple zeros b 1 , ... , bm. Let Bk be the coefficient of 1 / ( z -bk)
in the decomposition of R(z) into simple fractions. If C is a closed
contour such that bk rJ. C* (k = 1, ... ,m), show that
j R(z)dz=27rifBknc(bk)
c k=l
(1)