What is ((1+i)(1-i)^{-8})/((-9+3isqrt(3))^{-2)} ?

The solution to ((1+i)(1-i)^{-8})/((-9+3isqrt(3))^{-2)} is (27((1+sqrt(3))+(-sqrt(3)+1)i))/8

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simplify ((1+i)(1-i)^{-8})/((-9+3isqrt(3))^{-2)}

simplify

\frac{( 1 + i ) ( 1 - i ) ^{- 8}}{( - 9 + 3 i 3 ) ^{- 2}}

\frac{27 ( ( 1 + 3 ) + ( - 3 + 1 ) i )}{8}

Solution steps

\frac{( 1 + i ) ( 1 - i ) ^{- 8}}{( - 9 + 3 i 3 ) ^{- 2}}

Simplify

(1 + i) (1 - i)^{- 8} : \frac{1 + i}{( 1 - i ) ^{8}}

= \frac{\frac{1 + i}{( 1 - i ) ^{8}}}{( - 9 + 3 i 3 ) ^{- 2}}

Apply the fraction rule:

\frac{\frac{a}{b}}{c} = \frac{a}{b \cdot c}

= \frac{1 + i}{( 1 - i ) ^{8} ( - 9 + 3 i 3 ) ^{- 2}}

Simplify

(1 - i)^{8} (- 9 + 3 i 3)^{- 2} : \frac{( 1 - i ) ^{8}}{- 54 3 i + 54}

= \frac{1 + i}{\frac{( 1 - i ) ^{8}}{- 54 3 i + 54}}

Apply the fraction rule:

\frac{a}{\frac{b}{c}} = \frac{a \cdot c}{b}

= \frac{( 1 + i ) ( - 54 3 i + 54 )}{( 1 - i ) ^{8}}

Cancel

\frac{( 1 + i ) ( - 54 3 i + 54 )}{( 1 - i ) ^{8}} : \frac{( ( 1 + 3 ) + ( - 3 + 1 ) i ) \cdot 27}{8}

= \frac{( ( 1 + 3 ) + ( - 3 + 1 ) i ) \cdot 27}{8}

= \frac{27 ( ( 1 + 3 ) + ( - 3 + 1 ) i )}{8}

∣ x - 1 ∣ - 4 \leq 5

0 \leq x + 2 (4 x - 1) \leq 6

20 + 17 x = - 6 + 15 x

- 12 - 4 b = 4 - 2 b

(- 8)^{3} \cdot (- 8)^{3}

What is ((1+i)(1-i)^{-8})/((-9+3isqrt(3))^{-2)} ?
The solution to ((1+i)(1-i)^{-8})/((-9+3isqrt(3))^{-2)} is (27((1+sqrt(3))+(-sqrt(3)+1)i))/8