Question

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