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Chapter 2: Problem 8

Solve the equation $$ z^{4}=5(z-1)\left(z^{2}-z+1\right) $$

Short Answer

Expert verified
Question: Find the zeros of the quartic polynomial equation \(z^4 = 5(z-1)(z^2-z+1)\). Answer: Due to the complexity of the given polynomial equation, the solutions can be approximated using numerical methods such as Newton's method, the Bisection method, or the Secant method, or by using technology like a graphing calculator or computer software to find the zeros of the polynomial. Simple rational roots could not be found using the Rational Root Theorem.

Step by step solution

01

Simplify the product on the right-hand side

To do this, we need to multiply the two factors \((z-1)\) and \((z^2 - z + 1)\). We will use the distributive property of multiplication to accomplish this: \begin{align*} (z-1)(z^2-z+1) &= z(z^2-z+1) - 1(z^2-z+1) \\ &= (z^3 - z^2 + z) - (z^2 - z + 1) \\ &= z^3 - z^2 + z - z^2 + z - 1 \\ &= z^3 - 2z^2 + 2z - 1 \end{align*} Now, we have the simplified expression \(5(z^3 - 2z^2 + 2z - 1)\).
02

Write the equation in standard form

We will now substitute the simplified expression into the original equation. Then, we can rewrite the equation in standard form by bringing all the terms to the left-hand side: $$ z^4 = 5(z^3 - 2z^2 + 2z - 1) \\ z^4 = 5z^3 - 10z^2 + 10z - 5 \\ 0 = 5z^3 - 10z^2 + 10z - 5 - z^{4} $$ Rearrange the terms to get the standard form: $$ 0 = -z^4 + 5z^3 - 10z^2 + 10z - 5 $$
03

Look for possible factoring/solution finding methods

As the given polynomial equation is of degree 4, it does not factor easily. Therefore, we will resort to a numerical method called the Rational Root Theorem, which might help us find some solutions for this equation. The Rational Root Theorem states that if a rational number \(p/q\) is a root of the polynomial equation, then \(p\) is a factor of the constant term, and \(q\) is a factor of the leading coefficient. In our case, the constant term is 5, and the leading coefficient is -1. Factors of 5 are 卤1 and 卤5, and factors of -1 are 卤1. This suggests that the possible rational roots of the polynomial are: $$ \pm 1, \quad \pm 5 $$
04

Test possible rational roots

We will plug the possible rational roots into the polynomial equation to check if the equation holds true for any of them: For \(z=1\): $$ 0 = -(1)^4 + 5(1)^3 - 10(1)^2 + 10(1) - 5 \\ 0 = -1+5-10+10-5 \\ 0 \neq 1 $$ For \(z=-1\): $$ 0 = -(-1)^4 + 5(-1)^3 - 10(-1)^2 + 10(-1) - 5 \\ 0 = -1 -5 -10 -10 -5 \\ 0\neq -31 $$ For \(z=5\): $$ 0 = -(5)^4 + 5(5)^3 - 10(5)^2 + 10(5) - 5 \\ 0 = -625 + 625 - 250 + 50 - 5 \\ 0 \neq -255 $$ For \(z=-5\): $$ 0 = -(-5)^4 + 5(-5)^3 - 10(-5)^2 + 10(-5) - 5 \\ 0 = -625 - 625 - 250 - 50 - 5 \\ 0 \neq -1555 $$ It turns out that none of the possible rational roots are solutions to the equation.
05

Use a numerical method to approximate the solutions

Since we could not find any simple solutions, at this point, we should use a numerical method such as Newton's method, the Bisection method, or the Secant method, or rely on technology like a graphing calculator or a computer software to approximate the solutions. This will provide us with the values of z where the polynomial equation equals zero.

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Most popular questions from this chapter

Find polar representations for the following complex numbers: (a) \(z_{1}=6+6 i \sqrt{3}\); (b) \(z_{2}=-\frac{1}{4}+i \frac{\sqrt{3}}{4}\); (c) \(z_{3}=-\frac{1}{2}-i \frac{\sqrt{3}}{2}\); (d) \(z_{4}=9-9 i \sqrt{3}\); (e) \(z_{5}=3-2 i\); (f) \(z_{6}=-4 i\).

Let \(U_{n}=\left\\{\varepsilon_{0}, \varepsilon_{1}, \varepsilon_{2}, \ldots, \varepsilon_{n-1}\right\\} .\) Prove the following: (a) \(\varepsilon_{j} \cdot \varepsilon_{k} \in U_{n}\), for all \(j, k \in\\{0,1, \ldots, n-1\\}\) (b) \(\varepsilon_{j}^{-1} \in U_{n}\), for all \(j \in\\{0,1, \ldots, n-1\\}\).

For certain real values of \(a, b, c\), and \(d\), the equation $$ x^{4}+a x^{3}+b x^{2}+c x+d=0 $$ has four nonreal roots. The product of two of these roots is \(13+i\), and the sum of the other two roots is \(3+4 i\), where \(i=\sqrt{-1}\). Find \(b\).

Compute the following products using the polar representation of a complex number: (a) \(\left(\frac{1}{2}-i \frac{\sqrt{3}}{2}\right)(-3+3 i)(2 \sqrt{3}+2 i)\); (b) \((1+i)(-2-2 i) \cdot i\) (c) \(-2 i \cdot(-4+4 \sqrt{3} i) \cdot(3+3 i)\) (d) \(3 \cdot(1-i)(-5+5 i)\). Verify your results using the algebraic form.

Let \(z_{1}, z_{2}, z_{3}, z_{4}\) be the roots of \(\left(\frac{z-i}{2 z-i}\right)^{4}=1\). Determine the value of $$ \left(z_{1}^{2}+1\right)\left(z_{2}^{2}+1\right)\left(z_{3}^{2}+1\right)\left(z_{4}^{2}+1\right) . $$
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