What is 1/((1+sqrt(3)i)^4) ?

The solution to 1/((1+sqrt(3)i)^4) is -1/32+(sqrt(3))/(32)i

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simplify 1/((1+sqrt(3)i)^4)

simplify

\frac{1}{( 1 + 3 i ) ^{4}}

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

Solution steps

\frac{1}{( 1 + 3 i ) ^{4}}

Expand

(1 + 3 i)^{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

y (x^{2} + x y - 2 y^{2}) d x + x (3 y^{2})

0.6 (\frac{P + 2 b - 1.5 a}{2.5 a} + 1)

tan (2 x) (1 - tan^{2} (x))

x^{2} (1 + x)

\frac{3 x ( n ) - x ( n - 8 ) ( - 1 ) ^{k}}{2}

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