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

The solution to (-1+sqrt(3)i)^4 is -8+8sqrt(3)i

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

simplify

(- 1 + 3 i)^{4}

- 8 + 83 i

Solution steps

(- 1 + 3 i)^{4}

Apply binomial theorem:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = - 1, b = 3 i

= i = 0 \sum 4 (i 4) (- 1)^{(4 - i)} (3 i)^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} (- 1)^{4} (3 i)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (- 1)^{3} (3 i)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (- 1)^{2} (3 i)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (- 1)^{1} (3 i)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (- 1)^{0} (3 i)^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (- 1)^{4} (3 i)^{0} : 1

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (- 1)^{3} (3 i)^{1} : - 43 i

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (- 1)^{2} (3 i)^{2} : 18 i^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (- 1)^{1} (3 i)^{3} : - 123 i^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (- 1)^{0} (3 i)^{4} : 9 i^{4}

= 1 - 43 i + 18 i^{2} - 123 i^{3} + 9 i^{4}

Simplify

1 - 43 i + 18 i^{2} - 123 i^{3} + 9 i^{4} : 83 i - 8

= 83 i - 8

Rewrite in standard complex form:

- 8 + 83 i

= - 8 + 83 i

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

- 9 - \frac{( - 12 )}{3}

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(r^{3} + s^{3}) d r + s^{2} (3 r + k s) d s

x (x - 1)