Question:

If \( z = 2(\cos 60^\circ + i \sin 60^\circ) \), find the value of \( z^3 \).

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To compute powers of complex numbers in trigonometric form, use De Moivre's theorem. Verify carefully by converting to rectangular form if the result seems inconsistent with options.
Updated On: May 30, 2025
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The Correct Option is A

Approach Solution - 1

Given: \[ z = 2(\cos 60^\circ + i \sin 60^\circ) = 1 + i \sqrt{3} \] Using binomial expansion: \[ z^3 = (1 + i \sqrt{3})^3 \] First, \[ z^2 = (1 + i \sqrt{3})^2 = 1 + 2 i \sqrt{3} + (i \sqrt{3})^2 = 1 + 2 i \sqrt{3} - 3 = -2 + 2 i \sqrt{3} \] Then, \[ z^3 = z^2 \times z = (-2 + 2 i \sqrt{3})(1 + i \sqrt{3}) \] \[ = -2 - 2 i \sqrt{3} + 2 i \sqrt{3} + 2 (i \sqrt{3})^2 = -2 + 0 + 2(-3) = -2 - 6 = -8 \] Hence, \[ z^3 = -8 \]
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Approach Solution -2

To solve the problem, we need to find the value of $z^3$ given the complex number $z = 2(\cos 60^\circ + i \sin 60^\circ)$.

1. Expressing $z$ in Polar Form: 
Given:
$ z = 2(\cos 60^\circ + i \sin 60^\circ) $
Here, the modulus $r = 2$ and the argument $\theta = 60^\circ$.

2. Using De Moivre's Theorem:
According to De Moivre's theorem:
$ z^n = r^n \left(\cos n\theta + i \sin n\theta \right) $
For $n = 3$, we have:
$ z^3 = 2^3 \left(\cos 3 \times 60^\circ + i \sin 3 \times 60^\circ \right) $

3. Calculating the Power and Angles:
$ r^3 = 2^3 = 8 $
$ 3 \times 60^\circ = 180^\circ $

4. Substituting Values:
$ z^3 = 8 \left(\cos 180^\circ + i \sin 180^\circ \right) $

5. Evaluating Trigonometric Functions:
$ \cos 180^\circ = -1 $
$ \sin 180^\circ = 0 $

6. Final Calculation:
$ z^3 = 8 \times (-1 + 0 \times i) = -8 $

Final Answer:
The value of $z^3$ is $ {-8} $.

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