Q48 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics

If $\mathrm{i}^2=-1$ then $(1+\sqrt{3} \mathrm{i})^{2022}-(\sqrt{3}-\mathrm{i})^{2022}=$

  • Correct answer
    A
    $2^{2023}$
  • B
    0
  • C
    $2^{2022}$
  • D
    $3^{1011}$

Solution

$\begin{aligned} & \text (1+\sqrt{3} i)^{2022}-(\sqrt{3}-i)^{2022} \\ & =2^{2022}\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{2022}-2^{2022}\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^{2022} \\ & =2^{2022}\left[\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)^{2022}-\left(\cos \frac{\pi}{6}-i \sin \frac{\pi}{6}\right)^{2022}\right] \\ & =2^{2022}\left[\left(e^{\left.i \frac{\pi}{3}\right)^{2022}}-\left(e^{\left.\left.-i \frac{\pi}{6}\right)^{2022}\right]}\right.\right.\right. \\ & =2^{2022}\left[e^{i \frac{2022 \pi}{6}}-e^{\left.-i \frac{2022 \pi}{6}\right]}\right] \\ & =2^{2022}\left[e^{i \cdot 674 \pi}-e^{-i \cdot 337 \pi}\right]=2^{2022}\left[e^{i \cdot 0}-e^{-i \pi}\right] \\ & =2^{2022}[1-(-1)]=2^{2022} \times 2=2^{2023} .\end{aligned}$

Hence, the correct answer is option: 'A'.

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