Finding the Value of \(x^n + \frac{1}{{{x^n}}}\) using De Moivre's Theorem for Complex Numbers

The problem asks us to find the value of the expression \(x^n + \frac{1}{{{x^n}}}\) given that \(x = \cos \theta + i \sin \theta\). This involves working with complex numbers in polar form and applying a fundamental theorem in complex analysis, De Moivre's Theorem. Understanding complex numbers and their properties is key to solving this.

Given:

• \(x = \cos \theta + i \sin \theta\)

We need to find the value of \(x^n + \frac{1}{{{x^n}}}\).

Applying De Moivre's Theorem

De Moivre's Theorem is very useful for finding powers of complex numbers in polar form. It states that for any real number \(n\) and complex number \(\cos \theta + i \sin \theta\), we have:

\((\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta)\)

Calculating \(x^n\)

Using De Moivre's Theorem directly on the given expression for \(x\):

\(x^n = (\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta)\)

This gives us the value of \(x^n\).

Calculating \(\frac{1}{{{x^n}}}\)

To find \(\frac{1}{{{x^n}}}\), we can write it as \(x^{-n}\). Applying De Moivre's Theorem with the power \(-n\):

\(\frac{1}{{{x^n}}} = x^{-n} = (\cos \theta + i \sin \theta)^{-n}\)

According to De Moivre's Theorem, this is equal to:

\(\cos(-n\theta) + i \sin(-n\theta)\)

Using the trigonometry identities \(\cos(-\alpha) = \cos(\alpha)\) and \(\sin(-\alpha) = -\sin(\alpha)\), we simplify this:

\(\frac{1}{{{x^n}}} = \cos(n\theta) - i \sin(n\theta)\)

Alternatively, we could take the reciprocal of \(x^n\) and rationalize, which also leads to the same result for this specific expression.

Adding \(x^n\) and \(\frac{1}{{{x^n}}}\)

Now we add the expressions we found for \(x^n\) and \(\frac{1}{{{x^n}}}\) to find the value of the required expression \(x^n + \frac{1}{{{x^n}}}\):

\(x^n + \frac{1}{{{x^n}}} = (\cos(n\theta) + i \sin(n\theta)) + (\cos(n\theta) - i \sin(n\theta))\)

Simplifying the Expression

Let's combine the real and imaginary parts of the expression:

\(x^n + \frac{1}{{{x^n}}} = \cos(n\theta) + \cos(n\theta) + i \sin(n\theta) - i \sin(n\theta)\)

The terms \(+i \sin(n\theta)\) and \(-i \sin(n\theta)\) cancel each other out.

\(x^n + \frac{1}{{{x^n}}} = 2 \cos(n\theta)\)

Conclusion on the Value

Thus, the value of the expression \(x^n + \frac{1}{{{x^n}}}\) is \(2 \cos(n\theta)\). This result is a direct consequence of applying De Moivre's Theorem to complex numbers in polar form. It highlights a useful identity when dealing with powers of complex numbers on the unit circle. This type of problem is common in mathematics and engineering mathematics, requiring understanding of trigonometry and complex number properties.