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)

i$\sqrt 3$

Answer 2

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

So, the correct answer is (C): $i\sqrt 3$