What is (1-isqrt(3))^{-4} ?

The solution to (1-isqrt(3))^{-4} is -1/32-(sqrt(3))/(32)i

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simplify (1-isqrt(3))^{-4}

simplify

(1 - i 3)^{- 4}

- \frac{1}{32} - \frac{3}{32} i

Solution steps

(1 - i 3)^{- 4}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

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

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

Expand

(1 - i 3)^{4} : 83 i - 8

= \frac{1}{8 3 i - 8}

Rewrite

83 i - 8

in standard complex form:

- 8 + 83 i

= \frac{1}{- 8 + 8 3 i}

Multiply by the conjugate

\frac{- 8 - 8 3 i}{- 8 - 8 3 i}

= \frac{1 \cdot ( - 8 - 8 3 i )}{( - 8 + 8 3 i ) ( - 8 - 8 3 i )}

Simplify

\frac{1 \cdot ( - 8 - 8 3 i )}{( - 8 + 8 3 i ) ( - 8 - 8 3 i )} : \frac{- 8 - 8 3 i}{256}

= \frac{- 8 - 8 3 i}{256}

Apply the fraction rule:

\frac{a + bi}{c} = \frac{a}{c} + \frac{b}{c} i

= \frac{- 8}{256} + \frac{- 8 3}{256} i

Simplify

\frac{- 8}{256} + \frac{- 8 3}{256} i : - \frac{1}{32} - \frac{3}{32} i

= - \frac{1}{32} - \frac{3}{32} i

\frac{4 x ^{2}}{2}

(\frac{2}{s} - \frac{1}{s ^{3}})^{2}

\frac{( 2 ^{4 x + 4} )}{( 2 ^{x + 1} )}

i^{2021}

16 \frac{1}{2}

What is (1-isqrt(3))^{-4} ?
The solution to (1-isqrt(3))^{-4} is -1/32-(sqrt(3))/(32)i