Question

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

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

Answer 1

Answer: (C)

If in a complex number $a+ib$, the ratio $a:b$ is $1:\sqrt{3}$
then it always convert the complex number in $\omega$
Since, $\omega =-\frac{1}{2}+\frac{\sqrt{3}}{2}i$
$\therefore 4+5(-\frac{1}{2}+\frac{i\sqrt{3}}{2}{)}^{334}+3(-\frac{1}{2}+\frac{i\sqrt{3}}{2}{)}^{365}$
$=4+5{\omega }^{334}+3{\omega }^{365}$
$=4+5.({\omega }^3{)}^{111}.\omega +3.({\omega }^3{)}^{121}.{\omega }^2$
$=4+5\omega +3{\omega }^2[\because {\omega }^3=1]$
$=1+3+2\omega +3(1+\omega +{\omega }^2)=1+2\omega +3\times 0$
$[\because 1+\omega +{\omega }^2=0]$
$=1+(-1+\sqrt{3}i)=\sqrt{3}i$