What is (1-isqrt(3))^{-2} ?

The solution to (1-isqrt(3))^{-2} is -1/8+(sqrt(3))/8 i

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simplify (1-isqrt(3))^{-2}

simplify

(1 - i 3)^{- 2}

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

Solution steps

(1 - i 3)^{- 2}

Apply exponent rule:

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

(1 - i 3)^{- 2} = \frac{1}{( 1 - i 3 ) ^{2}}

= \frac{1}{( 1 - i 3 ) ^{2}}

Expand

(1 - i 3)^{2} : - 23 i - 2

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

Rewrite

- 23 i - 2

in standard complex form:

- 2 - 23 i

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

Multiply by the conjugate

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

= \frac{1 \cdot ( - 2 + 2 3 i )}{( - 2 - 2 3 i ) ( - 2 + 2 3 i )}

Simplify

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

= \frac{- 2 + 2 3 i}{16}

Apply the fraction rule:

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

= \frac{- 2}{16} + \frac{2 3}{16} i

Simplify

\frac{- 2}{16} + \frac{2 3}{16} i : - \frac{1}{8} + \frac{3}{8} i

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

(2 x - 12)^{2}

\frac{- x}{x ^{2}}

6 (x + 2)

\frac{12 a ^{4}}{3 ( a ^{5} ) ^{3}}

\frac{5 x}{x + 3} + \frac{15}{x + 3}

What is (1-isqrt(3))^{-2} ?
The solution to (1-isqrt(3))^{-2} is -1/8+(sqrt(3))/8 i