• # question_answer If $i=\sqrt{-1},$ then $4+5{{\left( -\frac{1}{2}+\frac{i\sqrt{3}}{2} \right)}^{334}}$   $+3{{\left( -\frac{1}{2}+\frac{i\sqrt{3}}{2} \right)}^{365}}$is equal to [IIT 1999] A) $1-i\sqrt{3}$ B) $-1+i\sqrt{3}$ C) $i\sqrt{3}$ D) $-i\sqrt{3}$

Given equation is $4+5{{\left( -\frac{1}{2}+i\frac{\sqrt{3}}{2} \right)}^{334}}+3{{\left( -\frac{1}{2}+i\frac{\sqrt{3}}{2} \right)}^{365}}$ $=4+5{{\left( \cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3} \right)}^{334}}$$+3{{\left( \cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3} \right)}^{365}}$ $=4+5\left[ \cos \frac{668}{3}\pi +i\sin \frac{668}{3}\pi \right]$$3\left[ \cos \frac{730}{3}\pi +i\sin \frac{730}{3}\pi \right]$ $=4+5\left[ \cos \left( 222\pi +\frac{2\pi }{3} \right)+i\sin \left( 222\pi +\frac{2\pi }{3} \right) \right]$$+3\left[ \cos \left( 243\pi +\frac{\pi }{3} \right)+i\sin \left( 243\pi +\frac{\pi }{3} \right) \right]$ $=4+5\left( \cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3} \right)+3\left( -\cos \frac{\pi }{3}-i\sin \frac{\pi }{3} \right)$ $=4+5\left( -\frac{1}{2}+i\frac{\sqrt{3}}{2} \right)+3\left( -\frac{1}{2}-i\frac{\sqrt{3}}{2} \right)$ $=4-4+2i\frac{\sqrt{3}}{2}=i\sqrt{3}$.