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

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

We have,
$4+5{(\frac{-1}{2}+\frac{i\sqrt{3}}{2})}^{334}+3{(\frac{-1}{2}+\frac{i\sqrt{3}}{2})}^{365}$
$=4+5(\omega {)}^{334}+3(\omega {)}^{365}$
where $\omega =e^{i2\pi /3}=\frac{-1}{2}+\frac{i\sqrt{3}}{2}$
$\omega =$cube root of unity
$[\because {\omega }^3=1]$
$[\because 1+\omega +{\omega }^2=0]$
$=4+5\omega (\omega {)}^{333}+3{\omega }^2({\omega }^{363})$
$=4+5\omega +3{\omega }^2$
$1+2\omega +(3+3\omega +3{\omega }^2)$
$=1+2(\frac{-1}{2}+\frac{i\sqrt{3}}{2})$
$=i\sqrt{3}$