228 CHAPTER 6 • COMPLEX INTECRATION
- Show that fc z-^1 dz = 27ri, where C is the square with vertices 1 ± i and -1 ± i
and having positive orientation. - Show that fct<OJ (4z^2 - 4z + 5)-
1
dz = 0. - Find fc (z^2 - zr^1 dz for
(a) circle c =ct ( 1) = {z: lz - 11 = 2} having positive orientation.
(b) circle c =ct (1) = { z : lz -l l = ~} having positive orientation.- Find J 0 (2z - 1) (z^2 - z)-
1
dz for the
(a) circle c =ct (0) = {z: lzl = 2} having positive orientation.
(b) circle c =ct (0) = { z : Jzl = ~} having positive orientation.- Let C be the triangle with vertices 0, 1, and i and having positive orientation.
Parametrize C and show that
(a) fc 1 dz= 0.
(b) f 0 z dz= 0.- Evaluate fc (4z^2 + 4z - 3r^1 dz= fc (2z - 1)-^1 (2z+3)-^1 dz for
(a) the circle c =ct (0).
(b) t he circle C =Ci(-~)= {z: lz+ ~I= l}.
(c) the circle C = Ct (O).- Use Green's theorem to show that the area enclosed by a simple closed contour C
is ~ f c x dy - y dx. - Parametrize Ci (0) with z (t) = cost+ i sin t, for -7r $ t $ 7r. Use the principal
branch of the square root function: z ~ = r; cos £ + ir; sin £, for -7r < (} $ ,. , to
find fct(o) z; dz. Hint: Take limits as t--> -7r.
10. Evaluate fc (z^2 - 1)-
1
dz for the contours shown in Figure 6.29.y y(a) (b)Figw-e 6.29