- 7 • THE ARGUMENT PRINCIPLE AND RoucHii:'s THEOREM 335
Solution Let f (z) = 7z + l, then f (z) + g (z) = z^4. At points on the circle
C1 (0) = {z: lzl = 1} we have the relationI/ (z) + 9 (z)I = jz^41 = 1 < 6 = 171-111 ~ l7z - I I = I/ (z)I.
The function f has one zero at the point-~ in the disk D1 (0), and the hypothe-
ses of Corollary 8.2 hold on the circle C 1 (0). Therefore, g has one zero inside
D1 (0).
-------.... EXERCISES FOR SECTION 8.7
I. Let f (z) = z^5 - z. Find the number of times the image f (C) winds around the
origin if(a) c = c! (0).
(b) C is t he rectangle with vertices ± ~ ± 3i.
(c) C = Cz (0).
(d) C = C1.2s (i).- Show that four of the five roots of t he equation z^5 + 15z + 1 = 0 belong to the
annulus A(~,2,0) = {z:! < lzl < 2}.
- Let g (z) = z^5 + 4z - 15.
(a) Show that there are no zeros in D1 (O).
(b) Show that there are five zeros in Dz (0). Hint: Consider f (z) = -z^5.
Remark: A factorization of the polynomial using numerical approxi-
mations for the coefficients is(z - 1.546) ( z^2 - 1.340z + 2.857) (z^2 + 2.885z + 3.3 9 7).- Let g (z) = z^3 + 9z+ 27.
(a) Show that there are no zeros in D2 (O).
(b) Show that there are three zeros in D4 (0).
Remark: A factorization of t he polynomial using numerical approxi-
mations for the coefficients is(z + 2.047) (z^2 - 2.047z + 13 .19).- Let g (z) = z^5 + 6z^2 + 2z + 1.
(a) Show that there are two zeros in D 1 (0).
(b) Show that there are five zeros in D2 (0).