What is (1+isqrt(3))^{-10} ?

The solution to (1+isqrt(3))^{-10} is -1/2048+(sqrt(3))/(2048)i

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simplify (1+isqrt(3))^{-10}

simplify

(1 + i 3)^{- 10}

- \frac{1}{2048} + \frac{3}{2048} i

Solution steps

(1 + i 3)^{- 10}

Apply exponent rule:

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

(1 + i 3)^{- 10} = \frac{1}{( 1 + i 3 ) ^{10}}

= \frac{1}{( 1 + i 3 ) ^{10}}

Expand

(1 + i 3)^{10} : - 5123 i - 512

= \frac{1}{- 512 3 i - 512}

Rewrite

- 5123 i - 512

in standard complex form:

- 512 - 5123 i

= \frac{1}{- 512 - 512 3 i}

Multiply by the conjugate

\frac{- 512 + 512 3 i}{- 512 + 512 3 i}

= \frac{1 \cdot ( - 512 + 512 3 i )}{( - 512 - 512 3 i ) ( - 512 + 512 3 i )}

Simplify

\frac{1 \cdot ( - 512 + 512 3 i )}{( - 512 - 512 3 i ) ( - 512 + 512 3 i )} : \frac{- 512 + 512 3 i}{1048576}

= \frac{- 512 + 512 3 i}{1048576}

Apply the fraction rule:

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

= \frac{- 512}{1048576} + \frac{512 3}{1048576} i

Simplify

\frac{- 512}{1048576} + \frac{512 3}{1048576} i : - \frac{1}{2048} + \frac{3}{2048} i

= - \frac{1}{2048} + \frac{3}{2048} i

\frac{x}{8} > 4

28 x^{2} \leq 0

- 3 x + 16 = - 4 (4 + 3 x) + 5 x

n^{2} + 9 n + 20

\frac{x ^{3}}{x ^{8}}

What is (1+isqrt(3))^{-10} ?
The solution to (1+isqrt(3))^{-10} is -1/2048+(sqrt(3))/(2048)i