Page 88 - Calculus with Complex Numbers

Ffgure 8.4

where

hl g(.::)
z) = 1 + fl .
z)

The condition Iflz) I> Ig (z) Iimplies that h (z) must satisfy the inequality

I/l (z) - 1 I< 1 for all z eEy . lt follows that h (z) cannot circulate the origin as z

goes round y (Figure 8.4). Therefore by the argument principle we must have
h'(z) ''Z ''-' ''
h
b
z
/
1 ç
as required.
Ayplication W e canuse Rouché's theorem to show for example that the zeros
of the polynomial plz) = .:3 + .:2 + 3 a1l lie in the annulus 1 < IzI < 2. lf we take
f U) = 2S, . % U) = 22 + 3, then for Izl= 2 we have

Therefore by Rouché's theorem plz) = f (z) + g (z) and f (z) have the same
number of zeros inside Iz I = 2. But f (z) = .:3 has 3 zeros inside I zl = 2 in the

form of a triple zero at z = 0. Hence also plz) has 3 zeros inside IzI= 2.
lf instead we take flz) EEE 3, g (z) = .:3 + z2, then for I z I= 1 we have

Therefore by Rouché's theorem p (z) = flz) + g (z) and flz) EEE 3 have the same

number of zeros inside I z I= 1. But f (z) has no zeros inside Iz I= 1. Hence

neither has p (z) (Figure 8.5).