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Dominik [7]
3 years ago
15

De moirve's (√3-i ÷ √3+i)^6 = 1

Mathematics
1 answer:
bulgar [2K]3 years ago
7 0

(√3 - <em>i </em>) / (√3 + <em>i</em> ) × (√3 - <em>i</em> ) / (√3 - <em>i</em> ) = (√3 - <em>i</em> )² / ((√3)² - <em>i</em> ²)

… = ((√3)² - 2√3 <em>i</em> + <em>i</em> ²) / (3 - <em>i</em> ²)

… = (3 - 2√3 <em>i</em> - 1) / (3 - (-1))

… = (2 - 2√3 <em>i</em> ) / 4

… = 1/2 - √3/2 <em>i</em>

… = √((1/2)² + (-√3/2)²) exp(<em>i</em> arctan((-√3/2)/(1/2))

… = exp(<em>i</em> arctan(-√3))

… = exp(-<em>i</em> arctan(√3))

… = exp(-<em>iπ</em>/3)

By DeMoivre's theorem,

[(√3 - <em>i </em>) / (√3 + <em>i</em> )]⁶ = exp(-6<em>iπ</em>/3) = exp(-2<em>iπ</em>) = 1

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