Evaluating Complex Expression with Cube Roots of Unity

The given expression is \({\left( {\frac{{ - 1 + {\rm{i}}\sqrt 3 }}{2}} \right)^{\rm{n}}} + {\left( {\frac{{ - 1 - {\rm{i}}\sqrt 3 }}{2}} \right)^{\rm{n}}}\), where \(n\) is an integer that is not a multiple of 3, and \({\rm{i}} = \sqrt { - 1}\).

We recognize the terms inside the parentheses as special complex numbers related to the cube roots of unity. Let \(\omega\) be one of the complex cube roots of unity.

• The complex number \(\frac{{ - 1 + {\rm{i}}\sqrt 3 }}{2}\) is one of the complex cube roots of unity, commonly denoted by \(\omega\).
• The complex number \(\frac{{ - 1 - {\rm{i}}\sqrt 3 }}{2}\) is the other complex cube root of unity, which is equal to \(\omega^2\).

The fundamental properties of the cube roots of unity (1, \(\omega\), \(\omega^2\)) are:

• \(\omega^3 = 1\)
• \(1 + \omega + \omega^2 = 0\)

Using these properties, we can rewrite the given expression in terms of \(\omega\):

Expression \( = \left( \frac{-1 + i\sqrt{3}}{2} \right)^n + \left( \frac{-1 - i\sqrt{3}}{2} \right)^n = \omega^n + (\omega^2)^n = \omega^n + \omega^{2n}\).

We are given that \(n\) is an integer but not a multiple of 3. This means that when \(n\) is divided by 3, the remainder is either 1 or 2.

Let's consider the two possible cases for \(n\):

Case 1: \(n\) is of the form \(3k+1\), where \(k\) is an integer.

In this case, the expression becomes:

\(\omega^n + \omega^{2n} = \omega^{3k+1} + \omega^{2(3k+1)} = \omega^{3k+1} + \omega^{6k+2}\)

Using the property \(\omega^3 = 1\), we have \(\omega^{3k} = (\omega^3)^k = 1^k = 1\) and \(\omega^{6k} = (\omega^3)^{2k} = 1^{2k} = 1\).

So, \(\omega^{3k+1} + \omega^{6k+2} = \omega^{3k} \cdot \omega^1 + \omega^{6k} \cdot \omega^2 = 1 \cdot \omega + 1 \cdot \omega^2 = \omega + \omega^2\).

From the property \(1 + \omega + \omega^2 = 0\), we know that \(\omega + \omega^2 = -1\).

Thus, for \(n = 3k+1\), the value of the expression is -1.

Case 2: \(n\) is of the form \(3k+2\), where \(k\) is an integer.

In this case, the expression becomes:

\(\omega^n + \omega^{2n} = \omega^{3k+2} + \omega^{2(3k+2)} = \omega^{3k+2} + \omega^{6k+4}\)

Using the property \(\omega^3 = 1\), we have \(\omega^{3k} = (\omega^3)^k = 1\) and \(\omega^{6k} = (\omega^3)^{2k} = 1\).

So, \(\omega^{3k+2} + \omega^{6k+4} = \omega^{3k} \cdot \omega^2 + \omega^{6k} \cdot \omega^4 = 1 \cdot \omega^2 + 1 \cdot (\omega^3 \cdot \omega^1) = \omega^2 + 1 \cdot (1 \cdot \omega) = \omega^2 + \omega\).

Again, from the property \(1 + \omega + \omega^2 = 0\), we know that \(\omega^2 + \omega = -1\).

Thus, for \(n = 3k+2\), the value of the expression is -1.

In both cases where \(n\) is not a multiple of 3, the value of the expression \({\left( {\frac{{ - 1 + {\rm{i}}\sqrt 3 }}{2}} \right)^{\rm{n}}} + {\left( {\frac{{ - 1 - {\rm{i}}\sqrt 3 }}{2}} \right)^{\rm{n}}}\) is -1.

The final answer is -1.

Revision Table: Cube Roots of Unity

Additional Information: Properties of Roots of Unity

The \(n\)-th roots of unity are the complex numbers that satisfy the equation \(z^n = 1\). These roots are given by the formula \(z_k = e^{i \frac{2\pi k}{n}} = \cos\left(\frac{2\pi k}{n}\right) + i \sin\left(\frac{2\pi k}{n}\right)\) for \(k = 0, 1, \dots, n-1\).

For \(n=3\), the cube roots of unity are:

• \(k=0\): \(z_0 = \cos(0) + i \sin(0) = 1\)
• \(k=1\): \(z_1 = \cos\left(\frac{2\pi}{3}\right) + i \sin\left(\frac{2\pi}{3}\right) = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\) (This is our \(\omega\))
• \(k=2\): \(z_2 = \cos\left(\frac{4\pi}{3}\right) + i \sin\left(\frac{4\pi}{3}\right) = -\frac{1}{2} - i\frac{\sqrt{3}}{2}\) (This is our \(\omega^2\))

The sum of the \(n\)-th roots of unity is always 0 for \(n > 1\). For \(n=3\), \(1 + \omega + \omega^2 = 0\).

Powers of \(\omega\) repeat in a cycle of 3:

• \(\omega^1 = \omega\)
• \(\omega^2 = \omega^2\)
• \(\omega^3 = 1\)
• \(\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega\)
• \(\omega^5 = \omega^3 \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2\)
• \(\omega^6 = (\omega^3)^2 = 1^2 = 1\)

In general, \(\omega^m = \omega^{m \pmod 3}\), where \(m \pmod 3\) is the remainder when \(m\) is divided by 3.

When \(n\) is not a multiple of 3, the remainder \(n \pmod 3\) is either 1 or 2. If \(n \equiv 1 \pmod 3\), then \(n = 3k+1\). \(\omega^n = \omega^{3k+1} = \omega\). Also \(2n = 6k+2\), so \(2n \equiv 2 \pmod 3\). \(\omega^{2n} = \omega^{6k+2} = \omega^2\). The sum is \(\omega + \omega^2 = -1\). If \(n \equiv 2 \pmod 3\), then \(n = 3k+2\). \(\omega^n = \omega^{3k+2} = \omega^2\). Also \(2n = 6k+4\), so \(2n \equiv 1 \pmod 3\). \(\omega^{2n} = \omega^{6k+4} = \omega\). The sum is \(\omega^2 + \omega = -1\).