1550251515-Classical_Complex_Analysis__Gonzalez_

Sequences, Series, and Special Functions 571

this particular case we can be much more specific about the number of its

zeros, its distribution, the nature of those zeros, and so on, and there is

an extensive literature {see [9] and [24]) about the properties of the zeros

of polynomials. For example, we have already seen that a polynomial of

degree n (n ~ 1) has exactly n zeros, and in Section 5.14 we have noted

that a polynomial has no zeros outside a circle of radius sufficiently large

with center at the origin. In what follows we propose to discuss a few

additional properties.

Theorem 8.25 Let P(z) be a polynomial of degree n ~ 1 with zeros a 1 ,

az, ... , an not necessarily distinct. Then

P'(z) _ n 1

P(z) - '2.: z -ak

k=l

for any z-:/:-ak (k = 1, ... ,n).

Proof From (5.14-3), namely,

P(z) = an(z - ai)(z - az) · · · (z - an)'

we have

(8.10-1)

P'(z) = an(z - az) · · · (z - an)+ an(z - ai)(z - aa) · · · (z - an)

+ · · · + an(z -ai)(z -az) · · · (z - an-1)

Hence for z -:/:-ak we obtain

P'(z) 1 1 1 n 1

P(z) = z - ai + z - az + ... + z - an = '2.: z - ak

k=l

Theorem 8.26 (Lucas, 1874). If all zeros of a polynomial P(z) lie in a

half-plane H, then all zeros of its derivative P'(z) lie in H.

Proof Suppose that the zeros ak of P(z) are in the right half-plane H

determined by the oriented line L: z = a+ bt, b -:/:- 0, t E JR. so that,

according to (2.4-15),

H = { z: Im z ~ a < 0}

Hence for all z EH'= C-H we have Im[(z-a)/b] ~ O, and it follows that

Im z - ak = Im (z -a) - (ak - a)

b b

z - a ak - a

=Im---Im-->0

b b