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

\frac{f}{f - g} - \frac{g}{g - f}

6 \times 8

4 x - 26

\frac{1}{cos ( x )} - cos (x)

\frac{sin ( x )}{sin ( x )}

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