What is the value of \({\left( {\frac{{ - 1 + i\sqrt 3 }}{2}} \right)^{3n}} + {\left( {\frac{{ - 1 - i\sqrt 3 }}{2}} \right)^{3n}}\) ? Where \(i = \sqrt { - 1} ?\)
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Understanding the Complex Numbers in the Expression
The given expression is \({\left( {\frac{{ - 1 + i\sqrt 3 }}{2}} \right)^{3n}} + {\left( {\frac{{ - 1 - i\sqrt 3 }}{2}} \right)^{3n}}\), where \(i = \sqrt { - 1}\).
Let's look at the complex numbers inside the parentheses:
- The first term is \( \frac{{ - 1 + i\sqrt 3 }}{2} \). This complex number is a well-known value related to the cube roots of unity.
- The second term is \( \frac{{ - 1 - i\sqrt 3 }}{2} \). This is the complex conjugate of the first term, also related to the cube roots of unity.
Identifying Cube Roots of Unity
The cube roots of unity are the solutions to the equation \( z^3 = 1 \). These solutions are:
- \( 1 \)
- \( \omega = \frac{{ - 1 + i\sqrt 3 }}{2} \)
- \( \omega^2 = \frac{{ - 1 - i\sqrt 3 }}{2} \)
Thus, the expression can be written in terms of \( \omega \) and \( \omega^2 \) as \( (\omega)^{3n} + (\omega^2)^{3n} \).
Applying Properties of Cube Roots of Unity
A key property of the non-real cube roots of unity (\( \omega \) and \( \omega^2 \)) is that \( \omega^3 = 1 \). Using this property, we can simplify the terms raised to the power \( 3n \).
- The first term is \( (\omega)^{3n} = (\omega^3)^n \). Since \( \omega^3 = 1 \), this becomes \( (1)^n \).
- The second term is \( (\omega^2)^{3n} = ((\omega^2)^3)^n \). We know that \( (\omega^2)^3 = \omega^6 = (\omega^3)^2 = 1^2 = 1 \). Alternatively, using the property \( \omega^3 = 1 \), we have \( (\omega^2)^{3n} = (\omega^{3n})^2 = ((\omega^3)^n)^2 = (1^n)^2 = 1^2 = 1 \). So, this term becomes \( (1)^n \).
For any integer value of \( n \), \( 1^n = 1 \).
Calculating the Final Value of the Expression
Now, we substitute the simplified terms back into the expression:
\[ {\left( {\frac{{ - 1 + i\sqrt 3 }}{2}} \right)^{3n}} + {\left( {\frac{{ - 1 - i\sqrt 3 }}{2}} \right)^{3n}} = (\omega)^{3n} + (\omega^2)^{3n} \] \[ = (\omega^3)^n + (\omega^6)^n \] \[ = (1)^n + (1)^n \] \[ = 1 + 1 \] \[ = 2 \]Therefore, the value of the given complex expression is 2.
Revision Table: Complex Number Concepts
| Concept | Description | Example |
|---|---|---|
| Complex Number | A number of the form \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i^2 = -1 \). | \( 3 + 4i \) |
| Cube Roots of Unity | Solutions to the equation \( z^3 = 1 \). | \( 1, \frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2} \) |
| Properties of Cube Roots of Unity | \( 1 + \omega + \omega^2 = 0 \), \( \omega^3 = 1 \). | If \( \omega = \frac{-1 + i\sqrt{3}}{2} \), then \( \omega^3 = 1 \). |
Additional Information: De Moivre's Theorem
Another way to approach this problem is by using the polar form of complex numbers and De Moivre's Theorem.
Let the complex number be \( z = r(\cos \theta + i \sin \theta) \). De Moivre's Theorem states that for any integer \( n \),
\[ z^n = r^n(\cos n\theta + i \sin n\theta) \]For the first term \( \frac{{ - 1 + i\sqrt 3 }}{2} \):
- Magnitude \( r = \left| \frac{-1 + i\sqrt{3}}{2} \right| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \).
- Argument \( \theta \): \( \cos \theta = -\frac{1}{2} \) and \( \sin \theta = \frac{\sqrt{3}}{2} \). This gives \( \theta = \frac{2\pi}{3} \) (or 120°).
So, \( \frac{{ - 1 + i\sqrt 3 }}{2} = 1 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right) \).
Applying De Moivre's Theorem for the power \( 3n \):
\[ \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right)^{3n} = \cos \left( 3n \cdot \frac{2\pi}{3} \right) + i \sin \left( 3n \cdot \frac{2\pi}{3} \right) \] \[ = \cos(2n\pi) + i \sin(2n\pi) \]Since \( n \) is an integer, \( \cos(2n\pi) = 1 \) and \( \sin(2n\pi) = 0 \). So, the first term evaluates to \( 1 \).
For the second term \( \frac{{ - 1 - i\sqrt 3 }}{2} \):
- Magnitude \( r = 1 \).
- Argument \( \phi \): \( \cos \phi = -\frac{1}{2} \) and \( \sin \phi = -\frac{\sqrt{3}}{2} \). This gives \( \phi = \frac{4\pi}{3} \) (or 240°).
So, \( \frac{{ - 1 - i\sqrt 3 }}{2} = 1 \left( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right) \).
Applying De Moivre's Theorem for the power \( 3n \):
\[ \left( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right)^{3n} = \cos \left( 3n \cdot \frac{4\pi}{3} \right) + i \sin \left( 3n \cdot \frac{4\pi}{3} \right) \] \[ = \cos(4n\pi) + i \sin(4n\pi) \]Since \( n \) is an integer, \( \cos(4n\pi) = 1 \) and \( \sin(4n\pi) = 0 \). So, the second term also evaluates to \( 1 \).
The sum is \( 1 + 1 = 2 \), which confirms the result obtained using cube roots of unity properties.
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