What is ((1-isqrt(3))/(-1+i))^4 ?

The solution to ((1-isqrt(3))/(-1+i))^4 is 2-2sqrt(3)i

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simplify ((1-isqrt(3))/(-1+i))^4

simplify

(\frac{1 - i 3}{- 1 + i})^{4}

2 - 23 i

Solution steps

(\frac{1 - i 3}{- 1 + i})^{4}

Simplify

\frac{1 - i 3}{- 1 + i} : \frac{- 1 - 3}{2} + \frac{- 1 + 3}{2} i

= \frac{- 1 - 3}{2} + \frac{- 1 + 3}{2} i

Simplify

(\frac{- 1 - 3}{2} + \frac{- 1 + 3}{2} i)^{4} : \frac{( - 1 - 3 + ( - 1 + 3 ) i ) ^{4}}{16}

= \frac{( - 1 - 3 + ( - 1 + 3 ) i ) ^{4}}{16}

Expand

(- 1 - 3 + (- 1 + 3) i)^{4} : - 323 i + 32

= \frac{- 32 3 i + 32}{16}

Factor out common term

32 : 32 (- 3 i + 1)

= \frac{32 ( - 3 i + 1 )}{16}

Divide the numbers:

\frac{32}{16} = 2

= 2 (- 3 i + 1)

Expand

2 (- 3 i + 1) : - 23 i + 2

= - 23 i + 2

Rewrite in standard complex form:

2 - 23 i

= 2 - 23 i

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What is ((1-isqrt(3))/(-1+i))^4 ?
The solution to ((1-isqrt(3))/(-1+i))^4 is 2-2sqrt(3)i