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Chapter 6: Q24E (page 332)

LetP(r1)=anr1n+an−1r1n−1+...+a1r1+a0 be a polynomialwith real coefficientsan,....,a0 . Prove that if r1 is azero ofP(r) , then so is its complex conjugate r1. [Hint:Show thatP(r)¯=P(r¯) , where the bar denotes complexconjugation.]

Short Answer

Expert verified

If r1 is a zero of the given polynomial P(r) then so is its complex conjugate r1.

Step by step solution

01

Step 1: Complex Conjugation

A complex conjugate of a complex number is another complex number that has the same real part as the original complex number and the imaginary part has the same magnitude but opposite sign. The product of a complex number and its complex conjugate is a real number.

02

Use of Complex Conjugation properties

Suppose complex number r1 is a zero of the given polynomial P(r). Hence P(r1)=0:

P(r1)=0⇒P(r1)¯=0¯=0

Using properties of complex conjugation:

P(r1)¯=(anr1n+an−1r1n−1+...+a1r1+a0)¯=anr1n¯+an−1r1n−1¯+...+a1r1¯+a0=P(r1¯)=0

Hence, the final answer is:

Ifr1 is a zero of the given polynomial P(r) then so is its complex conjugate r1.

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Most popular questions from this chapter

On a smooth horizontal surface, a mass of m1 kg isattached to a fixed wall by a spring with spring constantk1 N/m. Another mass of m2 kg is attached to thefirst object by a spring with spring constant k2 N/m. Theobjects are aligned horizontally so that the springs aretheir natural lengths. As we showed in Section 5.6, thiscoupled mass–spring system is governed by the systemof differential equations

m1d2xdt2+(k1+k2)x−k2y=0

m2d2ydt2−k2x+k2y=0

Let’s assume that m1 = m2 = 1, k1 = 3, and k2 = 2.If both objects are displaced 1 m to the right of theirequilibrium positions (compare Figure 5.26, page 283)and then released, determine the equations of motion forthe objects as follows:

(a)Show that x(2) satisfies the equationx(4)(t)+7x''(t)+6x(t)=0

(b) Find a general solution x(2) to (36).

(c) Substitute x(2) back into (34) to obtain a generalsolution for y(2)

(d) Use the initial conditions to determine the solutions,x(2) and y(2), which are the equations of motion.

Let y1x2= Cerx, where C (≠0) and r are real numbers,be a solution to a differential equation. Supposewe cannot determine r exactly but can only approximateit by r. Let (x) =Cerxand consider the error

|y(x)−y~(x)|

(a) If r andr~are positive, r ≠­ , show that the errorgrows exponentially large as x approaches + ∞.

(b) If r andrare negative, r≠ , show that the errorgoes to zero exponentially as x approaches + ∞.

use the method of undetermined coefficients to determine the form of a particular solution for the given equation.

y'''+y''-2y=xex+1

use the annihilator method to determinethe form of a particular solution for the given equation.

y''+2y'+2y=e-xcosx+x2

Find a general solution for the given linear system using the elimination method of Section 5.2.

d2xdt2−x+5y=02x+d2ydt2+2y=0
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