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Table 3.3.
Complex Zeros of Complex Polynomials 43
Procedure for Decreasing Roots by Amount h
1. Set Z;=h in (3.14) and find C_I usingk=n-I, ... ,O,
- 1; note the extra subscript added to (3.14).
2. Set b o = c_ l
; replace ai with Ci, i= n-1,... ,0; and
replace n with n-I,
3, Do steps I and 2 again, but equate b l =C_l'
4, Continue finding b 2 , b 3
,,,., b n through the n =0 cycle.
5, Find the roots of (3.37); then the roots of (3.12) are Zj = si + h.
in a more general application of synthetic division:
fez) = fez,) + (z- z,)(Co + c,z+ C,Z2). (3.40)
The first term on the right side of (3.40) is zero by definition if z, is a root of
fez); but (3.40) is valid for evaluating fez) for any z, not necessarily a root.
That first term is found as the value of c_ I
when (3.14) is calculated through
k = - I instead of just through k = 0 as previously applied. Suppose that z, = h,
and (3.36) is substituted for the linear term in (3.40). Then (3.38) is the result
of synthetic division cycle (3.14) on (3.12), and bo is obtained as C_ I when that
cycle is carried on through k=O. Now note that b, in (3.39) relates to (3.38) as
b o in (3.38) is related to (3.12). So synthetic division starting with co' c
"
and c 2
(found by the last synthetic division cycle) will yield b
"
co, and c, in (3.39).
The procedure in Table 3.3 finds a new polynomial, F(s), as in (3.37), given
polynomial fez), as in (3.12), so that s=z-h.
Example 3.6. Given polynomial f(z) in (3.33) with roots 98 + jO and 99 + jO,
find the corresponding polynomial F(s) having roots that are 100 less.
n=2, h= 100, a 2 =2, a, = -394. ao= 19404;
k= 1: c, =2+(100)XO=2
k=O: c o = -394+(100)x2= -194
k= - I: c_ I
= 19404+ (100) X(-194) =4= bo
n=l, h=100, a, =2, ao=-194;
k=O: c o =2+(100)XO=2
k=-l: c_ I
=-194+(100)X2=6=b,
n=O, h= 100, a o =2;
k= -I: c_ I
=2+(100)XO=2=b,
The polynomial with roots - 2 + jO and - I + jO is thus found to be
F(s)=4+6s+2s'.
(3.41 )
(3.42)