1. **Problem Statement:** Find the values of $n$ such that $\left(1 + \sqrt{3}i\right)^n$ is a real number. 2. **Formula and Important Rules:** We use the polar form of complex numbers. A complex number $z = x + yi$ can be written as $z = r(\cos \theta + i \sin \theta)$ where: - $r = \sqrt{x^2 + y^2}$ is the modulus, - $\theta = \tan^{-1}\left(\frac{y}{x}\right)$ is the argument. Raising to the power $n$: $$ z^n = r^n \left(\cos(n\theta) + i \sin(n\theta)\right) $$ For $z^n$ to be real, the imaginary part must be zero: $$ \sin(n\theta) = 0 $$ which implies $$ n\theta = k\pi, \quad k \in \mathbb{Z} $$ 3. **Step-by-step Solution:** - Given $z = 1 + \sqrt{3}i$, calculate modulus: $$ r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2 $$ - Calculate argument: $$ \theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} $$ - Express $z$ in polar form: $$ z = 2 \left(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}\right) $$ - Raise to power $n$: $$ z^n = 2^n \left(\cos \frac{n\pi}{3} + i \sin \frac{n\pi}{3}\right) $$ - For $z^n$ to be real, set imaginary part zero: $$ \sin \frac{n\pi}{3} = 0 $$ - This happens when: $$ \frac{n\pi}{3} = k\pi \implies n = 3k, \quad k \in \mathbb{Z} $$ 4. **Final Answer:** $$ \boxed{n = 3k, \quad k \in \mathbb{Z}} $$ This means $n$ must be any integer multiple of 3 for $\left(1 + \sqrt{3}i\right)^n$ to be real.