082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

9.2 Complex numbers 327
9.2.1 Thecomplexconjugate
Ifz =a +bj,wedefineitscomplexconjugatetobethenumberz =a −bj;thatis,we
change the sign of the imaginary part.
Example9.3 Write down the complex conjugates of
(a) −7+j (b) 6−5j (c) 6 (d) j
Solution Tofindthecomplexconjugatesofthegivennumberswechangethesignoftheimaginary
parts. Apurelyreal number has animaginarypart0.We find
(a) −7 −j (b) 6 +5j (c) 6,thereisno imaginarypart toalter (d) −j
We recall that the solution of the quadratic equation 2x 2 +2x +5 = 0 yielded the two
complexnumbers − 1 2 + 3 2 jand−1 2 − 3 2 j,andnotethattheseformacomplexconjugate
pair. This illustratesamore general result:
When the polynomial equationP(x) = 0 has real coefficients, any complex roots
will always occur incomplex conjugatepairs.
Consider the following example.
Example9.4 Show that the equationx 3 −7x 2 +19x −13 = 0 has a root atx = 1 and find the other
roots.
Solution IfweletP(x) =x 3 −7x 2 +19x −13,thenP(1) =1−7 +19 −13 =0sothatx =1
is a root. This means thatx −1 must be a factor ofP(x) and so we can expressP(x) in
the form
P(x)=x 3 −7x 2 +19x−13=(x−1)(αx 2 +βx+γ)
=αx 3 +(β−α)x 2 +(γ−β)x−γ
where α, β and γ arecoefficientstobedetermined.Comparingthecoefficientsofx 3 we
find α = 1. Comparing the constant coefficients we find γ = 13. Finally, comparing
coefficients ofxwefind β = −6,and hence
P(x) =x 3 −7x 2 +19x −13 = (x −1)(x 2 −6x +13)
Theothertwo rootsofP(x) = 0 arefoundby solvingthe quadraticequation
x 2 −6x+13=0,thatis
x = 6 ± √ 36−52
2
= 6 ± √ −16
2
=3±2j
and again we note that the complex roots occur as a complex conjugate pair. This illustratesthegeneralresultgiveninSection1.4thatannth-degreepolynomialhasnroots.