The value of 1 - i-31 - i-36 - 1 - i-31 - i-36 is

The correct option is A 2 1+i√( 3 ) =2 e ^ iπ #160; / 3 #160; 1 i√( 3 ) =2 e ^ iπ #160; / 3 #160; ( 1+i√( 3 ) #160; 1 i√( 3 ...

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Question

The value of ${(\frac{1 + i \sqrt{3}}{1 - i \sqrt{3}})}^{6} + {(\frac{1 - i \sqrt{3}}{1 + i \sqrt{3}})}^{6}$ is

2

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{(\frac{1 + i \sqrt{3}}{1 - i \sqrt{3}})}^{6} + {(\frac{1 - i \sqrt{3}}{1 + i \sqrt{3}})}^{6}

Evaluate each of the following:

$(a) (- 30) \div 10$

$(b) 50 \div (- 5)$

$(c) (- 36) \div (- 9)$

$(d) (- 49) \div 49$

$(e) 13 \div [- 2 + 1]$

$(f) 0 \div (- 12)$

$(g) (- 31) \div [(- 30) + (- 1)]$

$(h) [(- 36) \div 12] \div 3$

$(i) [- 6 + 5] \div [- 2 + 1]$

{(\frac{1 + i \sqrt{3}}{1 - i \sqrt{3}})}^{6} + {(\frac{1 - i \sqrt{3}}{1 + i \sqrt{3}})}^{6}

{(\frac{1 + i \sqrt{3}}{1 - i \sqrt{3}})}^{6} + {(\frac{1 - i \sqrt{3}}{1 + i \sqrt{3}})}^{6}

i = \sqrt{- 1}

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

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