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Chapter 7: Problem 44

Solve each equation. Express answers in the form \(a+b i\). $$x^{4}+81=0$$

Short Answer

Expert verified
The solutions are \(3e^{i\frac{\pi}{8}}, 3e^{i\frac{5\pi}{8}}, 3e^{i\frac{9\pi}{8}}, 3e^{i\frac{13\pi}{8}} \) or \(3(\cos(\frac{\pi}{8}) + i\sin(\frac{\pi}{8})), 3(\cos(\frac{5\pi}{8}) + i\sin(\frac{5\pi}{8})), 3(\cos(\frac{9\pi}{8}) + i\sin(\frac{9\pi}{8})), 3(\cos(\frac{13\pi}{8}) + i\sin(\frac{13\pi}{8})) \).

Step by step solution

01

Rewrite the equation

Start by rewriting the equation, moving 81 to the right side: \[ x^4 = -81 \]
02

Factor the right side

Notice that -81 can be expressed as -1 * 81. Since 81 is also 9 squared, write the equation as: \[ x^4 = - (9^2) \]
03

Simplify using complex numbers

Recognize that \( -1 \) can be written as \( i^2 \), where \( i \) is the imaginary unit \( i = \sqrt{-1} \). Therefore: \[ x^4 = -(9^2) = (9i)^2 \]
04

Take the fourth root of both sides

Taking the fourth root on both sides of the equation: \[ x = \pm \sqrt[4]{(9i)^2} \]
05

Solve the roots

To solve for \( x \), we need to find all fourth roots of \( 9i \). Using the polar form of complex numbers, express \( 9i \) as: \[ 9i = 9e^{i\frac{\pi}{2}} \] The fourth roots are given by: \[ x = 9^{1/2} e^{i(\frac{\pi}{8} + k\frac{2\pi}{4})} \], where \( k = 0, 1, 2, 3 \).
06

Find all roots

Now calculate all four roots individually by substituting \( k = 0, 1, 2, 3 \): \[ x_1 = \sqrt{9} e^{i\frac{\pi}{8}} = 3 e^{i\frac{\pi}{8}}\] \[x_2 = \sqrt{9} e^{i\frac{5\pi}{8}} = 3 e^{i\frac{5\pi}{8}}\] \[x_3 = \sqrt{9} e^{i\frac{9\pi}_{8}} = 3 e^{i\frac{9\pi}{8}}\] \[x_4 = \sqrt{9} e^{i\frac{13\pi}{8}} = 3 e^{i\frac{13\pi}{8}} \]
07

Simplify to rectangular form

Converting each result to rectangular form using \( e^{i\theta} = \cos\theta + i\sin\theta \):\[x_1 = 3(\cos(\frac{\pi}{8}) + i\sin(\frac{\pi}{8}))\] \[x_2 = 3(\cos(\frac{5\pi}{8}) + i\sin(\frac{5\pi}{8}))\] \[x_3 = 3(\cos(\frac{9\pi}{8}) + i\sin(\frac{9\pi}{8}))\] \[x_4 = 3(\cos(\frac{13\pi}{8}) + i\sin(\frac{13\pi}{8}))\]

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Key Concepts

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

solving polynomial equations
Solving polynomial equations is a fundamental concept in precalculus and higher-level math. It involves finding the values of the variable that make the polynomial equal to zero.
For example, in the given exercise, we start with the polynomial equation \(x^4 + 81 = 0\). The goal is to solve for \(x\).
Here are the steps to solve such equations:
  • Rewrite the equation: Move all terms to one side of the equation, typically setting it equal to zero.
  • Factor or simplify: Break down complex parts into simpler components. Identify patterns or common factors. Write complex numbers if necessary.
  • Extract roots: Use algebraic methods or formulas to find the roots. For higher-degree polynomials, you might need to use the quadratic formula or synthetic division.
  • Verify solutions: Substitute back into the original equation to ensure they satisfy the equation.

Understanding these steps makes solving polynomial equations straightforward and manageable.
imaginary unit
The imaginary unit, denoted as \(i\), is a key concept in complex numbers. It is defined as \(i = \sqrt{-1}\).
This means \(i^2 = -1\). Imaginary units help in extending the real number system to include solutions to equations that do not have real solutions.
In the exercise, \(x^4 = -81\), the \(-81\) can be simplified using the imaginary unit:
  • Recognize that \(-81\) is equivalent to \(-(9^2)\).
  • Since \(i^2 = -1\), we can rewrite \(-(9^2)\) as \((9i)^2\).

By using the imaginary unit, it becomes possible to simplify and solve complex polynomial equations.
complex roots
Complex roots are the solutions to polynomial equations that involve complex numbers. When solving equations like \(x^4 + 81 = 0\), we often find that the solutions are not real but complex.
To find these roots, we can use the polar form of complex numbers:
  • Convert the complex number into polar form. For instance, \(9i = 9e^{i\frac{\pi}{2}}\).
  • Use the formula for finding \(k^{th}\) roots: \(x = r^{1/n} e^{i(\theta/n + \frac{2k\pi}{n})}\). Here, \(r\) is the modulus, \(\theta\) is the argument, and \(n\) is the root degree.
  • Substitute \(k = 0, 1, 2, 3\) to get all four roots.

These steps ensure we find all complex roots of the polynomial equation. Expressing these in rectangular form makes them easier to understand and visualize.

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Most popular questions from this chapter

Find \(z_{1} z_{2}\) and \(z_{1} / z_{2}\) for each pair of complex numbers, using trigonometric form. Write the answer in the form \(a+b i\). $$z_{1}=1+4 i, z_{2}=-4-2 i$$

Given that \(\mathbf{A}=\langle 3,1\rangle\) and \(\mathbf{B}=\langle- 2,3\rangle,\) find the magnitude and direction angle for each of the following vectors. $$\mathbf{A}-\mathbf{B}$$

A tall building casts a shadow of length 230 feet when the angle of elevation of the sun is \(48^{\circ} .\) Find the height of the building.

Simplify the expression \(\frac{1-\sin ^{2} x \csc ^{2} x+\sin ^{2} x}{\cos ^{2} x}\)

Show that the reciprocal of \(z=r(\cos \theta+i \sin \theta)\) is \(z^{-1}=r^{-1}(\cos \theta-i \sin \theta),\) provided \(r \neq 0\)
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