Answers to Exercises; Chapter (^5 329)
a few sketches, we can see that D2 U 03 (the image of the left semi-circle)
must wind around the origin n times. But then the left semi-circle must
surround all n zeros of the polynomial, leaving none for the right semi-circle
to surround.
3.12. (a) The Nyquist diagram indicates that the polynomial is unstable.
2 - 8i
(b) Every real zero must be negative. Suppose u + iv is a nonreal zero.
Then (u3 + 2u2 + 3u + 1) = v2(3u + 2) and 3u2 + 4u + 3 = v2. Eliminating
v2 leads to 8u3 + 16u2 + 14u + 5 = 0, whence u cannot be nonnegative.
Hence any nonreal zero must have negative real part.
(c) Every real zero must be negative. If u + vi is a nonreal zero, then
(u4 + 3u3 + u2 + u + 8) - (6u2 + 9u + l)v2 + v4 = 0
(4u3 + 9u2 + 2u + 1) = (4u + 3)v2
Eliminating v2 leads to
... (1)
... (2)
+ (4213 + 92 + 2u + 1)2 = 0.The leading coefficjent of the left side is negative and the constant coef-
ficient is positive. Hence the equation is satisfied for at least one positive
value of u. Such a positive value of u leads to a pair of real values of v, so