Q110 / 180 · COMEDK 2022 · Mathematics
$(i+\sqrt{3})^{100}+(i-\sqrt{3})^{100}+2^{100}$ is equal to
- Correct answerA0
- B1
- C-1
- D$2^{101}$
Solution
Let $z_1 = i + \sqrt{3} = 2(\cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6}) = 2e^{i\pi/6}$.
Then $z_1^{100} = 2^{100} e^{i 100\pi/6} = 2^{100} e^{i 50\pi/3} = 2^{100} e^{i (16\pi + 2\pi/3)} = 2^{100} e^{i 2\pi/3}$.
Using Euler's formula, $z_1^{100} = 2^{100} (\cos \dfrac{2\pi}{3} + i \sin \dfrac{2\pi}{3}) = 2^{100} (-\dfrac{1}{2} + i \dfrac{\sqrt{3}}{2})$.
Let $z_2 = i - \sqrt{3} = 2(\cos \dfrac{5\pi}{6} + i \sin \dfrac{5\pi}{6}) = 2e^{i 5\pi/6}$.
Then $z_2^{100} = 2^{100} e^{i 500\pi/6} = 2^{100} e^{i 250\pi/3} = 2^{100} e^{i (82\pi + 4\pi/3)} = 2^{100} e^{i 4\pi/3}$.
Using Euler's formula, $z_2^{100} = 2^{100} (\cos \dfrac{4\pi}{3} + i \sin \dfrac{4\pi}{3}) = 2^{100} (-\dfrac{1}{2} - i \dfrac{\sqrt{3}}{2})$.
Summing the terms: $z_1^{100} + z_2^{100} = 2^{100} (-\dfrac{1}{2} + i \dfrac{\sqrt{3}}{2} - \dfrac{1}{2} - i \dfrac{\sqrt{3}}{2}) = 2^{100} (-1) = -2^{100}$.
Adding $2^{100}$ to the sum: $-2^{100} + 2^{100} = 0$.
Answer: 0
Then $z_1^{100} = 2^{100} e^{i 100\pi/6} = 2^{100} e^{i 50\pi/3} = 2^{100} e^{i (16\pi + 2\pi/3)} = 2^{100} e^{i 2\pi/3}$.
Using Euler's formula, $z_1^{100} = 2^{100} (\cos \dfrac{2\pi}{3} + i \sin \dfrac{2\pi}{3}) = 2^{100} (-\dfrac{1}{2} + i \dfrac{\sqrt{3}}{2})$.
Let $z_2 = i - \sqrt{3} = 2(\cos \dfrac{5\pi}{6} + i \sin \dfrac{5\pi}{6}) = 2e^{i 5\pi/6}$.
Then $z_2^{100} = 2^{100} e^{i 500\pi/6} = 2^{100} e^{i 250\pi/3} = 2^{100} e^{i (82\pi + 4\pi/3)} = 2^{100} e^{i 4\pi/3}$.
Using Euler's formula, $z_2^{100} = 2^{100} (\cos \dfrac{4\pi}{3} + i \sin \dfrac{4\pi}{3}) = 2^{100} (-\dfrac{1}{2} - i \dfrac{\sqrt{3}}{2})$.
Summing the terms: $z_1^{100} + z_2^{100} = 2^{100} (-\dfrac{1}{2} + i \dfrac{\sqrt{3}}{2} - \dfrac{1}{2} - i \dfrac{\sqrt{3}}{2}) = 2^{100} (-1) = -2^{100}$.
Adding $2^{100}$ to the sum: $-2^{100} + 2^{100} = 0$.
Answer: 0
Hence, the correct answer is option: 'A'.