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} $.
Let \( z \) satisfy \( |z| = 1, \ z = 1 - \overline{z} \text{ and } \operatorname{Im}(z)>0 \)
Then consider:
Statement-I: \( z \) is a real number
Statement-II: Principal argument of \( z \) is \( \dfrac{\pi}{3} \)
Then: