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Use De Moivre's theorem to verify the solution given for each polynomial equation. $$2 z^{4}+3 z^{3}-4 z^{2}+2 z+12=0 ; z=1-i$$

Short Answer

Expert verified
z = 1 - i is a solution since using De Moivre's Theorem leads to zero when substituted into the polynomial.

Step by step solution

01

Rewrite the Given Complex Number in Polar Form

The given root is \( z = 1 - i \). First, express this in polar form. The magnitude \( r \) is calculated as \( r = \sqrt{1^2 + (-1)^2} = \sqrt{2} \). The argument \( \theta \) is \( \tan^{-1} \left( \frac{-1}{1} \right) = -\frac{\pi}{4} \). So, \( z = \sqrt{2} (\cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4})) \).
02

Apply De Moivre's Theorem to Compute Powers of z

De Moivre's Theorem states \( (r(\cos \theta + i \sin \theta))^n = r^n (\cos(n\theta) + i \sin(n\theta)) \). To calculate \( z^n \), find each required power of \( z \): \( z^2, z^3, \) and \( z^4 \).
03

Compute z^2

Using De Moivre’s Theorem, \( z^2 = (\sqrt{2})^2 \left( \cos(2 \cdot -\frac{\pi}{4}) + i \sin(2 \cdot -\frac{\pi}{4}) \right) = 2(\cos(-\frac{\pi}{2}) + i \sin(-\frac{\pi}{2})) = 2 \times 0 + 2i(-1) = -2i \).
04

Compute z^3

For \( z^3 \), \( z^3 = (\sqrt{2})^3 \left( \cos(3 \cdot -\frac{\pi}{4}) + i \sin(3 \cdot -\frac{\pi}{4}) \right) = 2\sqrt{2}(\cos(-\frac{3\pi}{4}) + i \sin(-\frac{3\pi}{4})) = 2\sqrt{2} (-\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}) = -2 - 2i \).
05

Compute z^4

For \( z^4 \), \( z^4 = (\sqrt{2})^4 \left( \cos(4 \cdot -\frac{\pi}{4}) + i \sin(4 \cdot -\frac{\pi}{4}) \right) = 4 (\cos(-\pi) + i\sin(-\pi)) = 4(-1 + 0i) = -4 \).
06

Substitute Back into the Polynomial Equation

Substitute \( z \), \( z^2 \), \( z^3 \), and \( z^4 \) into the polynomial equation \( 2z^4 + 3z^3 - 4z^2 + 2z + 12 = 0 \): \( 2(-4) + 3(-2 - 2i) - 4(-2i) + 2(1-i) + 12 \).
07

Perform the Algebraic Simplification

Calculate each term separately: \( 2(-4) = -8 \), \( 3(-2 - 2i) = -6 - 6i \), \( -4(-2i) = 8i \), \( 2(1-i) = 2 - 2i \), and \( 12 \). Combine all terms: \(-8 - 6 - 6i + 8i + 2 - 2i + 12 = 0 \).
08

Verify Both Real and Imaginary Parts Equal Zero

Combine like terms for real and imaginary components separately: \( -8 - 6 + 2 + 12 = 0 \) and \( -6i + 8i - 2i = 0 \). Both real and imaginary parts equal zero, which confirms \( z = 1 - i \) is a solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polar Form
Polar form is a way to express complex numbers using a combination of a magnitude and an angle. This is especially useful when dealing with multiplication and powers of complex numbers. For a complex number given as \( z = a + bi \), where \( a \) is the real part and \( b \) is the imaginary part, the polar form is \( z = r(\cos \theta + i \sin \theta) \). Here, \( r \) is the magnitude of the complex number, determined by \( r = \sqrt{a^2 + b^2} \).
\( \theta \) is the argument, or angle, which can be found using \( \tan^{-1} \left( \frac{b}{a} \right) \). This angle is usually in radians. Using polar form often makes operations like exponentiation and multiplication more straightforward, leveraging the properties of angles and magnitudes rather than traditional algebraic methods.
Complex Numbers
Complex numbers are numbers that have both a real part and an imaginary part. They are written in the form \( z = a + bi \), where \( a \) is the real part and \( b \) is the imaginary part, with \( i \) being the square root of \(-1\).
These numbers are not just theoretical; they have practical applications in engineering, physics, and computer science. Complex numbers allow for the representation of oscillations and waves, which are foundational in many areas of technology.
  • Adding complex numbers involves adding their real and imaginary components separately.
  • Multiplying them requires using the distributive property and the fact that \( i^2 = -1 \).
This leads to a very rich set of operations that extend the real number system into a broader set that includes both geometric and trigonometric interpretations.
Polynomial Equations
Polynomial equations are mathematical expressions equating a polynomial to zero. They have the general form \( a_nz^n + a_{n-1}z^{n-1} + \cdots + a_1z + a_0 = 0 \).
These equations can have multiple solutions or roots, which can be real or complex numbers. The degree of the polynomial, defined by the highest power of the variable \( z \), suggests the maximum number of solutions an equation might have.
  • For example, a quadratic equation \( az^2 + bz + c = 0 \) has two potential solutions.
  • A cubic one, like \( az^3+ bz^2 + cz + d = 0 \), offers up to three solutions.
Solving these often involves techniques such as factoring, using the quadratic formula, or applying De Moivre's Theorem when complex roots are involved.
Complex Roots
Complex roots refer to solutions of polynomial equations that are complex numbers. These roots are essential in understanding the behavior of polynomials, especially when real roots are not apparent.
Given a polynomial equation of degree \( n \), if it has complex roots, they often occur in conjugate pairs—if \( a + bi \) is a root, then \( a - bi \) is usually also a root.
  • Complex roots are acquired by extending real number solutions to accommodate the imaginary unit \( i \).
  • De Moivre's Theorem particularly aids in finding powers of these roots through polar coordinates.
By representing and manipulating in polar form, we leverage rotation and scaling, offering elegant solutions to understanding permutations of complex numbers in various equations.

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