Problem 31 - Contemporary Precalculus

Question

Solve the given equation in the complex number system. $$x^{6}=-1$$

Step-by-Step Solution

Answer

Question: Find the six complex roots of the equation \(x^6 = -1\). Answer: The six complex roots of the equation \(x^6 = -1\) are: 1. \(x_0 = \frac{\sqrt{3} + i}{2}\) 2. \(x_1 = -\frac{1 + i\sqrt{3}}{2}\) 3. \(x_2 = -\frac{\sqrt{3} - i}{2}\) 4. \(x_3 = \frac{-\sqrt{3} - i}{2}\) 5. \(x_4 = \frac{1 - i\sqrt{3}}{2}\) 6. \(x_5 = \frac{\sqrt{3} + i}{2}\)

Step 1: Rewrite the complex number in polar form

The given equation is $$x^{6}=-1$$ We can rewrite -1 as a complex number in polar form with \(\cos \theta = -1\) and \(\sin \theta = 0\). So, -1 can be expressed as $$-1 = r(\cos\theta + i\sin\theta) = r(\cos(\pi) + i\sin(\pi))$$ where \(r = 1\) and \(\theta = \pi\).

Step 2: Find the roots using De Moivre's theorem

According to De Moivre's theorem, $$x^n = r^n(\cos(n\theta) + i\sin(n\theta))$$ Here, we want to find the six roots of the equation \(x^6 = 1(\cos(\pi) + i\sin(\pi))\). To find these roots, we can write: $$x^{6} = 1^{\frac{6}{6}}(\cos(\frac{6\pi}{6}) + i\sin(\frac{6\pi}{6}))$$ So, $$x_k = \sqrt[6]{1}(\cos(\frac{\pi+2k\pi}{6}) + i\sin(\frac{\pi+2k\pi}{6}))$$ where \(k=0,1,2,...,5\).

Step 3: Calculate the roots

Compute the roots using the above formula for each value of k. For \(k=0\): $$x_0 = \sqrt[6]{1}(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}))= \frac{\sqrt{3} + i}{2}$$ For \(k=1\): $$x_1 = \sqrt[6]{1}(\cos(\frac{3\pi}{6}) + i\sin(\frac{3\pi}{6})) = -\frac{1 + i\sqrt{3}}{2}$$ For \(k=2\): $$x_2 = \sqrt[6]{1}(\cos(\frac{5\pi}{6}) + i\sin(\frac{5\pi}{6})) = -\frac{\sqrt{3} - i}{2}$$ For \(k=3\): $$x_3 = \sqrt[6]{1}(\cos(\frac{7\pi}{6}) + i\sin(\frac{7\pi}{6})) = \frac{-\sqrt{3} - i}{2}$$ For \(k=4\): $$x_4 = \sqrt[6]{1}(\cos(\frac{9\pi}{6}) + i\sin(\frac{9\pi}{6})) = \frac{1 - i\sqrt{3}}{2}$$ For \(k=5\): $$x_5 = \sqrt[6]{1}(\cos(\frac{11\pi}{6}) + i\sin(\frac{11\pi}{6})) = \frac{\sqrt{3} + i}{2}$$ These are the six complex roots of the given equation \(x^6 = -1\).

Key Concepts

Roots of Unity

Every complex number equation can have several solutions, known as roots. In the case of the equation \(x^6 = -1\), we are searching for six numbers, known as the sixth roots, that can be multiplied together to give minus one. These are a special type of complex numbers called the 'roots of unity'.
Roots of unity are complex numbers that when raised to a certain power, result in one. In general, if you have an equation \(x^n = 1\), the solution will be \(n\) distinct roots. However, in this exercise, we're dealing with \(x^6 = -1\), requiring a slight adjustment in thinking.
To obtain roots of such an equation, one typically uses the polar form of complex numbers. You'll follow a process using the equation:

• Convert the equation into polar coordinates.
• Use the known polar form expression of \(-1\) that is \(\pi\) in argument with \(r=1\).

These six unique solutions, known as complex sixth roots of \(-1\), create a symmetrical shape when plotted on the complex plane, often represented as vertices of a polygon.

De Moivre's Theorem

De Moivre's theorem is a powerful tool in complex numbers, especially helpful when dealing with powers and roots. It states that for a complex number expressed in polar form as \(r(\cos\theta + i\sin\theta)\), its \(n\)-th power can be calculated as \(r^n(\cos(n\theta) + i\sin(n\theta))\).
In the exercise where we have \(x^6 = -1\), De Moivre's theorem assists us in finding the roots of the equation by considering \(-1\) as \(1\cdot(\cos\pi + i\sin\pi)\).
When applying the theorem:

• Start by rewriting \( -1 \) in polar form.
• Express the given power \(x^6\) using De Moivre's formula.
• Find \(x_k\) using the expression \(\sqrt[6]{1}(\cos(\frac{\pi+2k\pi}{6}) + i\sin(\frac{\pi+2k\pi}{6}))\), for \(k=0\) to \(5\).

Through this process, you'll compute six different angles that correspond to the needed solutions. Each angle leads you to one of the complex roots. This seamless approach provides an elegant way to visualize and solve complex number equations.

Polar Coordinates

Polar coordinates are a way to express complex numbers that focus on their magnitude and direction, representing them as \(r(\cos\theta + i\sin\theta)\).
This method is especially useful when manipulating powers and roots of complex numbers. In the problem \(x^6 = -1\), representing \(-1\) in polar form makes finding its roots much more straightforward.
Here's how it works:

• Any complex number \(a + bi\) can be represented as \(r(\cos\theta + i\sin\theta)\), where \(r\) is the magnitude, calculated as \(\sqrt{a^2 + b^2}\), and \(\theta\) is the argument, given by \(\tan^{-1}(b/a)\).
• For \(-1\), we have \(r = 1\), and \(\theta = \pi\).
• This tells us that \(-1\) is located on the negative x-axis \(\pi\) radians around the unit circle, originating from the positive x-axis.

By converting complex numbers into polar form, De Moivre's theorem can be more neatly applied, simplifying the otherwise complex calculations. This approach supports understanding and solving equations involving complex numbers efficiently.