Question

If $(\sqrt{3} - i)^{50} = 2^{48}(x - iy)$, then $x^{2} + y^{2}$ equals

• A. 2
• B. 4
• C. 8
• D. 16

Hint:

Polar form shortcut: $(\sqrt{3}-i)^n = 2^n(\cos(n\pi/6) - i\sin(n\pi/6))$ (using angle $-\pi/6$). Reduce the angle modulo $2\pi$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Convert to polar form and use De Moivre's theorem.
Step 2: Detailed Explanation:
$\sqrt{3}-i = 2e^{-i\pi/6}$, so $(\sqrt{3}-i)^{50} = 2^{50}e^{-i\cdot 50\pi/6} = 2^{50}e^{-i25\pi/3}$.
$25\pi/3 = 8\pi + \pi/3$, so $e^{-i25\pi/3} = e^{-i\pi/3} = \dfrac{1}{2} - i\dfrac{\sqrt{3}}{2}$.
$(\sqrt{3}-i)^{50} = 2^{50}(\tfrac{1}{2}-i\tfrac{\sqrt{3}}{2}) = 2^{49}(1-i\sqrt{3}) = 2^{48}(2-2i\sqrt{3})$.
So $x=2$, $y=2\sqrt{3}$, giving $x^2+y^2 = 4+12=16$.
Step 3: Final Answer:
$x^2 + y^2 = 16$.