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Chapter 3: Problem 179

The roots of the equation \(z^{4}+4 z^{2}+16=0\) are (A) \(\pm 1+i \sqrt{3}\) (B) \(\pm 1-i \sqrt{3}\) (C) \(\pm 2+2 \sqrt{3} i\) (D) \(\pm 2-2 \sqrt{3} i\)

Short Answer

Expert verified
The roots are \(\pm 2 + 2\sqrt{3}i\) (option C).

Step by step solution

01

Use Substitution to Simplify

Start by letting \(w = z^2\). This substitution transforms the original equation \(z^4 + 4z^2 + 16 = 0\) into a quadratic equation: \(w^2 + 4w + 16 = 0\).
02

Solve the Quadratic Equation

Solve the quadratic equation \(w^2 + 4w + 16 = 0\) using the quadratic formula: \[w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a = 1\), \(b = 4\), and \(c = 16\). Substitute these values into the formula:
03

Calculate the Discriminant

Compute the discriminant \(b^2 - 4ac\): \(4^2 - 4 \cdot 1 \cdot 16 = 16 - 64 = -48\). The negative discriminant indicates that the roots are complex.
04

Find the Complex Roots

Using the quadratic formula, find the roots: \[w = \frac{-4 \pm \sqrt{-48}}{2} = \frac{-4 \pm i \sqrt{48}}{2}\]Simplify \(\sqrt{48}\) to \(4\sqrt{3}\), giving: \[w = -2 \pm 2i\sqrt{3}\].
05

Convert Back to Original Variable

The solutions \(w = -2 + 2i\sqrt{3}\) and \(w = -2 - 2i\sqrt{3}\) imply \(z^2 = -2 + 2i\sqrt{3}\) and \(z^2 = -2 - 2i\sqrt{3}\). Find \(z\) by taking the square root of these expressions.
06

Calculate Roots of Original Equation

The solutions for \(z^2 = -2 + 2i\sqrt{3}\) are \(z = \pm (\sqrt{-2 + 2i\sqrt{3}})\), and for \(z^2 = -2 - 2i\sqrt{3}\) are \(z = \pm (\sqrt{-2 - 2i\sqrt{3}})\). Evaluating these square roots gives the roots: \(z = \pm (2 + 2i\sqrt{3})\).
07

Match the Roots with the Given Options

Comparing the roots \(z = \pm (2 + 2i\sqrt{3})\) from the solution with the options provided in the exercise leads to the choice corresponding to (C) \(\pm 2 + 2\sqrt{3}i\).

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Key Concepts

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

Quadratic Equations
A quadratic equation is a polynomial equation of degree 2, characterized by the general form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). In the context of our original exercise, we encountered a higher-degree polynomial: \( z^4 + 4z^2 + 16 = 0 \). To simplify this problem, we used a substitution technique by letting \( w = z^2 \). This transformed our high-degree equation into the quadratic equation \( w^2 + 4w + 16 = 0 \).

Here are some key points about quadratic equations:
  • They typically have two solutions, known as "roots." These roots can be real or complex.
  • Quadratic equations can often be solved using the quadratic formula, factoring, or completing the square method.
  • The solutions represent the x-values where the parabola, defined by the quadratic equation, intersects the x-axis.
By simplifying higher-degree equations into quadratic ones, we make the task of solving them more approachable, using well-known methods and techniques.
Complex Numbers
Complex numbers extend the concept of regular numbers by including 'imaginary' components. A complex number is usually expressed in the form \( a + bi \), where \( a \) is the real part, \( b \) is the imaginary part, and \( i \) is the square root of \( -1 \). In our example, the quadratic equation \( w^2 + 4w + 16 = 0 \) had a negative discriminant, indicating complex roots:

  • Complex numbers are crucial because they provide solutions to equations that do not have real number solutions. For instance, find the roots of \( z^2 = -2 \cdot (1 - i\sqrt{3}) \).
  • The square root of a negative number is defined via the imaginary unit \( i \), as \( \sqrt{-1} = i \).
  • In mathematics, complex numbers have practical applications across fields like engineering, physics, and signal processing.

    Having a grasp of complex numbers enables us to solve equations with negative discriminants and understand the full range of possible solutions.
Discriminant Analysis
The discriminant is a part of the quadratic formula that helps determine the nature of the roots of a quadratic equation. For an equation \( ax^2 + bx + c = 0 \), the discriminant, \( \Delta \), is given by \( b^2 - 4ac \). The value of the discriminant reveals much about the type of roots we can expect:

  • If \( \Delta > 0 \), the equation has two distinct real roots.
  • If \( \Delta = 0 \), the equation has exactly one real root (also called a repeated or double root).
  • If \( \Delta < 0 \), the equation has two complex conjugate roots.
In our original exercise, we computed the discriminant \( b^2 - 4ac \) for the equation \( w^2 + 4w + 16 = 0 \) as \( 16 - 64 = -48 \). The negative value here indicated the presence of complex roots. By recognizing this, we were able to proceed to find those complex solutions accurately.

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Most popular questions from this chapter

If \(z\) and \(\omega\) are two non-zero complex numbers such that \(|z \omega|=1\), and \(\operatorname{Arg}(z)-\operatorname{Arg}(\omega)=\frac{\pi}{2}\), then \(\bar{Z} \omega\) is equal to \([\mathbf{2 0 0 3}]\) (A) 1 (B) \(-1\) (C) \(i\) (D) \(-i\)

If \(z\) is a complex number such that \(|z| \geq 2\), then the minimum value of \(\left|z+\frac{1}{2}\right|\) [2014] (A) is equal to \(\frac{5}{2}\) (B) lies in the interval \((1,2)\) (C) is strictly greater than \(\frac{5}{2}\) (D) is strictly greater than \(\frac{3}{2}\) but less than \(\frac{5}{2}\)

The roots of the equation \(z^{4}+1=0\) are (A) \((\pm 1 \pm i)\) (B) \((\pm 2 \pm 2 i)\) (C) \(\frac{1}{\sqrt{2}}(\pm 1 \pm i)\) (D) None of these

If \(i=\sqrt{-1}\), then \(4+5\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)^{334}+3\left(-\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)^{365}\) is equal to (A) \(1-i \sqrt{3}\) (B) \(-1+i \sqrt{3}\) (C) \(i \sqrt{3}\) (D) \(-\sqrt{3} i\)

The value of \(\sum_{k=1}^{10}\left(\sin \frac{2 k \pi}{11}+i \cos \frac{2 k \pi}{11}\right)\) is (A) \(\vec{i}\) (B) 1 (C) \(-1\) (D) \(-i\)
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