Cube roots of unity properties and location on the Argand plane

Cube Roots of Unity: Properties and Argand Plane Location Explore the properties of cube roots of unity and their geometric representation on the Argand plane, crucial for solving complex number problems in JEE.

Concept Overview

This question tests the understanding of the properties of the cube roots of unity, specifically $\omega$ and $\omega^2$, and their geometric interpretation on the Argand plane. Key properties include $\omega^3 = 1$, $1 + \omega + \omega^2 = 0$, and the fact that these roots, along with 1, form an equilateral triangle on the complex plane. This knowledge is fundamental for simplifying expressions involving roots of unity and solving various complex number problems.

Step 1: Define the cube roots of unity. The cube roots of unity are the solutions to the equation $z^3 = 1$. We can write 1 in polar form as $1 = e^{i(2k\pi)}$, where $k$ is an integer. $z^3 = e^{i(2k\pi)}$ Taking the cube root of both sides, we get: $z = (e^{i(2k\pi)})^{1/3} = e^{i(2k\pi/3)}$ For $k=0, 1, 2$, we get the distinct cube roots of unity.

Step 2: Calculate the distinct cube roots of unity. For $k=0$: $z_0 = e^{i(0)} = \cos(0) + i\sin(0) = 1$ For $k=1$: $z_1 = e^{i(2\pi/3)} = \cos(2\pi/3) + i\sin(2\pi/3) = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ Let's denote this root as $\omega$. For $k=2$: $z_2 = e^{i(4\pi/3)} = \cos(4\pi/3) + i\sin(4\pi/3) = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$ Notice that $z_2 = \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)^2 = \omega^2$. So, the three cube roots of unity are $1, \omega, \omega^2$.

Step 3: Verify the property $\omega^3 = 1$. Using the polar form of $\omega$: $\omega^3 = (e^{i(2\pi/3)})^3 = e^{i(2\pi)} = \cos(2\pi) + i\sin(2\pi) = 1$ This confirms that $\omega$ is indeed a cube root of unity.

Step 4: Verify the property $1 + \omega + \omega^2 = 0$. Substitute the values of $1, \omega, \omega^2$: $1 + \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) + \left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)$ $= 1 - \frac{1}{2} + i\frac{\sqrt{3}}{2} - \frac{1}{2} - i\frac{\sqrt{3}}{2}$ $= 1 - 1 + 0i = 0$ This property is crucial for simplifying many expressions.

Step 5: Geometric interpretation on the Argand plane. The cube roots of unity can be represented as points on the Argand plane. The root $1$ is at $(1, 0)$. The root $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ is at $(-\frac{1}{2}, \frac{\sqrt{3}}{2})$. The root $\omega^2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$ is at $(-\frac{1}{2}, -\frac{\sqrt{3}}{2})$. These three points lie on the unit circle (since their modulus is 1) and are equally spaced, forming the vertices of an equilateral triangle. The angle between consecutive roots from the origin is $2\pi/3$ radians or 120 degrees.

Step 6: Properties derived from the geometric interpretation. Since $1, \omega, \omega^2$ form an equilateral triangle centered at the origin, the sum of the vectors from the origin to these points is zero. This is consistent with $1 + \omega + \omega^2 = 0$. Also, the distance between any two roots is equal, confirming the equilateral nature of the triangle. For example, the distance between $1$ and $\omega$ is $|1 - \omega| = |1 - (-\frac{1}{2} + i\frac{\sqrt{3}}{2})| = |\frac{3}{2} - i\frac{\sqrt{3}}{2}| = \sqrt{(\frac{3}{2})^2 + (-\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{9}{4} + \frac{3}{4}} = \sqrt{\frac{12}{4}} = \sqrt{3}$.

Key Takeaways:

• The three cube roots of unity are $1, \omega, \omega^2$, where $\omega = e^{2\pi i/3}$.
• Fundamental properties are $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.
• Geometrically, the cube roots of unity form an equilateral triangle inscribed in the unit circle on the Argand plane, with vertices at $1, \omega, \omega^2$.
• These properties are essential for simplifying complex number expressions and solving problems involving roots of unity.

Answer: The cube roots of unity are $1, -\frac{1}{2} + i\frac{\sqrt{3}}{2}, -\frac{1}{2} - i\frac{\sqrt{3}}{2}$. They satisfy $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$, and form an equilateral triangle on the Argand plane.