Question

If $i=\sqrt{-1}$, then $4+5{(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)}^{334}+3{(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)}^{365}$ is equal to

• A. $1-i\sqrt{3}$
• B. $-1+i\sqrt{3}$
• C. $i\sqrt{3}$
• D. $-i\sqrt{3}$

Answer 1

The correct option is C $i\sqrt{3}$

$\omega =-\frac{1}{2}+\frac{i\sqrt{3}}{2}\left[cuberootofunity\right]\therefore 4+5{(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)}^{334}+3{(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)}^{365}=4+5{\omega }^{334}+3{\omega }^{365}=4+5.({\omega }^3{)}^{111}.\omega +3.({\omega }^3{)}^{121}.{\omega }^2=4+5\omega +3{\omega }^2[\because {\omega }^3=1]=1+3+2\omega +3\omega +3{\omega }^2=1+2\omega +3(1+\omega +{\omega }^2)=1+2\omega +3\times 0[\because 1+\omega +{\omega }^2=0]=1+(-1+\sqrt{3}i)=\sqrt{3}i$