Question

$(i+\sqrt{3}{)}^{100}+(i-\sqrt{3}{)}^{100}+2^{100}$ is equal to

• A. 0
• B. 1
• C. -1
• D. $2^{101}$

Answer 1

The correct option is A 0

$(i+\sqrt{3}{)}^{100}=1\ast (i+\sqrt{3}{)}^{100}=i^{100}(i+\sqrt{3}{)}^{100}$ (As $i^{4m}=1$ Here $m=25\therefore i^{100}=1$)

$=(i^2+i\sqrt{3}{)}^{100}=(-1+i\sqrt{3}{)}^{100}=(2\omega {)}^{100}$

Similarly $(i-\sqrt{3}{)}^{100}=(-1-\sqrt{3}{)}^{100}=(2{\omega }^2{)}^{100}$

$(i+\sqrt{3}{)}^{100}+(i-\sqrt{3}{)}^{100}+2^{100}=(2\omega {)}^{100}+(2{\omega }^2{)}^{100}+2^{100}=2^{100}({\omega }^{100}+{\omega }^{200}+1)$

$=2^{100}\left(0\right)=0$ ($1+{\omega }^m+{\omega }^{2m}=0ifm\ne 3n$)

$\therefore$ $(i+\sqrt{3}{)}^{100}+(i-\sqrt{3}{)}^{100}+2^{100}=0$