Q100 / 180 · COMEDK 2021 · Mathematics
If $(\sqrt{3}+i)^{100}=2^{99}(a+i b)$, then $a^{2}+b^{2}$ is equal to
- A$\sqrt{2}$
- Correct answerB4
- C$\sqrt{3}$
- DNone of these
Solution
Given the expression $(\sqrt{3} + i)^{100} = 2^{99}(a + ib)$.
Express the complex number $z = \sqrt{3} + i$ in polar form. The modulus is $r = |\sqrt{3} + i| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2$.
The argument $\theta$ is given by $\tan \theta = \dfrac{1}{\sqrt{3}}$, so $\theta = \dfrac{\pi}{6}$. Thus, $\sqrt{3} + i = 2(\cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6})$.
Using De Moivre's Theorem, $(\sqrt{3} + i)^{100} = [2(\cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6})]^{100} = 2^{100}(\cos \dfrac{100\pi}{6} + i \sin \dfrac{100\pi}{6})$.
Simplifying the angle, $\dfrac{100\pi}{6} = \dfrac{50\pi}{3} = 16\pi + \dfrac{2\pi}{3}$.
Therefore, $(\sqrt{3} + i)^{100} = 2^{100}(\cos \dfrac{2\pi}{3} + i \sin \dfrac{2\pi}{3}) = 2^{100}(-\dfrac{1}{2} + i \dfrac{\sqrt{3}}{2}) = 2^{99}(-1 + i\sqrt{3})$.
Comparing this with $2^{99}(a + ib)$, we get $a = -1$ and $b = \sqrt{3}$.
Then $a^2 + b^2 = (-1)^2 + (\sqrt{3})^2 = 1 + 3 = 4$.
Answer: 4
Express the complex number $z = \sqrt{3} + i$ in polar form. The modulus is $r = |\sqrt{3} + i| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2$.
The argument $\theta$ is given by $\tan \theta = \dfrac{1}{\sqrt{3}}$, so $\theta = \dfrac{\pi}{6}$. Thus, $\sqrt{3} + i = 2(\cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6})$.
Using De Moivre's Theorem, $(\sqrt{3} + i)^{100} = [2(\cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6})]^{100} = 2^{100}(\cos \dfrac{100\pi}{6} + i \sin \dfrac{100\pi}{6})$.
Simplifying the angle, $\dfrac{100\pi}{6} = \dfrac{50\pi}{3} = 16\pi + \dfrac{2\pi}{3}$.
Therefore, $(\sqrt{3} + i)^{100} = 2^{100}(\cos \dfrac{2\pi}{3} + i \sin \dfrac{2\pi}{3}) = 2^{100}(-\dfrac{1}{2} + i \dfrac{\sqrt{3}}{2}) = 2^{99}(-1 + i\sqrt{3})$.
Comparing this with $2^{99}(a + ib)$, we get $a = -1$ and $b = \sqrt{3}$.
Then $a^2 + b^2 = (-1)^2 + (\sqrt{3})^2 = 1 + 3 = 4$.
Answer: 4
Hence, the correct answer is option: 'B'.