If i=−1, then 4+5−12+i32334+3−12+i32365 is equal to

We know that ω=−12+32iThus, 4+5−12+i32334+3−12+i32365=4+5ω334+3ω365=4+5ω3111ω+3ω3123ω2=4+5(1)111ω+3(1)123ω2=4+5ω+3ω2=1+2ω+31+ω+ω2=1+2ω+3(0)=1+2ω=1−1+3i=3i

If i=−1, then 4+5−12+i32334+3−12+i32365 is equal to

We know that ω=−12+32iThus, 4+5−12+i32334+3−12+i32365=4+5ω334+3ω365=4+5ω3111ω+3ω3123ω2=4+5(1)111ω+3(1)123ω2=4+5ω+3ω2=1+2ω+31+ω+ω2=1+2ω+3(0)=1+2ω=1−1+3i=3i

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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

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$1 - i \sqrt{3}$

$- 1 + i \sqrt{3}$

$i \sqrt{3}$

$- \sqrt{3} i$

answer is C.

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Detailed Solution

We know that $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$

Thus, $4 + 5 {(- \frac{1}{2} + \frac{i \sqrt{3}}{2})}^{334} + 3 {(- \frac{1}{2} + \frac{i \sqrt{3}}{2})}^{365}$

$\begin{array}{l} = 4 + 5 ω^{334} + 3 ω^{365} \\ = 4 + 5 {(ω^{3})}^{111} ω + 3 {(ω^{3})}^{123} ω^{2} \\ = 4 + 5 (1)^{111} ω + 3 (1)^{123} ω^{2} \\ = 4 + 5 ω + 3 ω^{2} = 1 + 2 ω + 3 (1 + ω + ω^{2}) \\ = 1 + 2 ω + 3 (0) = 1 + 2 ω = 1 - 1 + \sqrt{3} i = \sqrt{3} i \end{array}$