Problem 89 - Precalculus Mathematics for Calculus

Question

Find the indicated roots, and graph the roots in the complex plane. The fourth roots of \(-1\)

Step-by-Step Solution

Answer

The four roots are \(\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}\), \(-\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}\), \(-\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\), and \(\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\).

Step 1: Understand the Problem

We need to find the four complex numbers such that when raised to the fourth power, they equal \(-1\). This can be written as finding the roots of the equation \(z^4 = -1\).

Step 2: Express the Number in Polar Form

The number \(-1\) can be expressed in polar form as \(1(\cos(\pi) + i\sin(\pi))\). Since any integer multiple of \(2\pi\) can be added to the angle, we also have \(-1 = 1(\cos(\pi + 2k\pi) + i\sin(\pi + 2k\pi))\), where \(k\) is an integer.

Step 3: Use the General Formula for Roots

The nth roots of a complex number given in polar form \(r(\cos(\theta) + i\sin(\theta))\) are \([r^{1/n} (\cos(\frac{\theta + 2k\pi}{n}) + i\sin(\frac{\theta + 2k\pi}{n})), k = 0, 1, ..., n-1]\). We apply this formula with \(n = 4\), \(r = 1\), and \(\theta = \pi\).

Step 4: Calculate the Four Roots

Substitute \(r = 1\), \(n = 4\), and \(\theta = \pi\) into the formula: \(z_k = \cos(\frac{\pi + 2k\pi}{4}) + i\sin(\frac{\pi + 2k\pi}{4})\) for \(k = 0, 1, 2, 3\). This gives:\(z_0 = \cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}\),\(z_1 = \cos(\frac{3\pi}{4}) + i\sin(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}\), \(z_2 = \cos(\frac{5\pi}{4}) + i\sin(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\),\(z_3 = \cos(\frac{7\pi}{4}) + i\sin(\frac{7\pi}{4}) = \frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\).

Step 5: Graph the Roots on the Complex Plane

Plot the four roots \(z_0, z_1, z_2, z_3\) on the complex plane. The points will form the vertices of a square centered at the origin with vertices on the unit circle. - \(z_0 = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}\) in the first quadrant,- \(z_1 = -\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}\) in the second quadrant,- \(z_2 = -\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\) in the third quadrant,- \(z_3 = \frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\) in the fourth quadrant.

Key Concepts

Polar Form

The concept of polar form is a way of expressing complex numbers using a magnitude and angle, rather than the typical real and imaginary components. This form is very useful when dealing with multiplication and roots of complex numbers.
Imagine a point on the complex plane. In polar form, this point can be represented as a distance (or magnitude) from the origin and an angle from the positive real axis.
A complex number can be written as:

• Cartesian form:
  • a + bi
• Polar form:
  • r( \(<\cos \theta + i\sin \theta\) )

Here, r is the distance from the origin (the modulus), and \( \theta \) is the angle which we call the argument.
For \( -1 \), a complex number in polar form becomes \( 1( \cos\(\pi\) + i\sin\(\pi\)\ ) \), where \( 1\) is the magnitude and \( \pi\) is the angle from the positive real axis.
This representation simplifies many complex number operations, particularly when finding roots.

Nth Roots

Finding the nth roots of a complex number is all about expressing the number in polar form first. This method allows us to break down the complexity into more manageable parts.
The formula to find the nth roots of a complex number in polar form is:

• \[ r^{1/n} \left( \cos \left( \frac{\theta + 2k\pi}{n} \right) + i\sin \left( \frac{\theta + 2k\pi}{n} \right) \right) \text{, for } k = 0, 1, 2, \ldots, n-1. \ \]

This formula uses the modulus \( r \) and the argument \( \theta \) to calculate all possible nth roots.
In our example, for the fourth roots of \( -1 \), we have \( n = 4 \), \( r = 1 \), and \( \theta = \pi \), leading to four distinct roots.
By plugging in the values and solving, we get four unique outcomes, which, when plotted, form a symmetrical pattern on the complex plane.
This illustrates the geometric beauty and symmetry of complex roots.

Complex Plane Graphing

Graphing complex numbers is a straightforward yet powerful way to visualize complex equations and their solutions. The complex plane consists of two axes:

• The real axis (horizontal)
• The imaginary axis (vertical)

Each complex number corresponds to a unique point on this plane.
For visualizing roots, after calculation using polar form, each root becomes a point on this plane.
In the case of the fourth roots of \( -1 \), plotting these roots yields an interesting geometric shape. The points: \( z_0 = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} \), \( z_1 = -\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} \), \( z_2 = -\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2} \), and \( z_3 = \frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2} \), form the vertices of a square centered at the origin.
This kind of plotting not only aids in understanding but also showcases the beauty of mathematical symmetry. By embracing both real and imaginary components, we unlock clearer insights into complex number behavior.