What is (-1+sqrt(3)i)^5 ?

The solution to (-1+sqrt(3)i)^5 is -16-16sqrt(3)i

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simplify (-1+sqrt(3)i)^5

simplify

(- 1 + 3 i)^{5}

- 16 - 163 i

Solution steps

(- 1 + 3 i)^{5}

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 5 (i 5) (- 1)^{(5 - i)} (3 i)^{i}

Expand summation

= \frac{5 !}{0 ! ( 5 - 0 ) !} (- 1)^{5} (3 i)^{0} + \frac{5 !}{1 ! ( 5 - 1 ) !} (- 1)^{4} (3 i)^{1} + \frac{5 !}{2 ! ( 5 - 2 ) !} (- 1)^{3} (3 i)^{2} + \frac{5 !}{3 ! ( 5 - 3 ) !} (- 1)^{2} (3 i)^{3} + \frac{5 !}{4 ! ( 5 - 4 ) !} (- 1)^{1} (3 i)^{4} + \frac{5 !}{5 ! ( 5 - 5 ) !} (- 1)^{0} (3 i)^{5}

Simplify

\frac{5 !}{0 ! ( 5 - 0 ) !} (- 1)^{5} (3 i)^{0} : - 1

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} (- 1)^{4} (3 i)^{1} : 53 i

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} (- 1)^{3} (3 i)^{2} : - 30 i^{2}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} (- 1)^{2} (3 i)^{3} : 303 i^{3}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} (- 1)^{1} (3 i)^{4} : - 45 i^{4}

Simplify

\frac{5 !}{5 ! ( 5 - 5 ) !} (- 1)^{0} (3 i)^{5} : 93 i^{5}

= - 1 + 53 i - 30 i^{2} + 303 i^{3} - 45 i^{4} + 93 i^{5}

Simplify

- 1 + 53 i - 30 i^{2} + 303 i^{3} - 45 i^{4} + 93 i^{5} : - 163 i - 16

= - 163 i - 16

Rewrite in standard complex form:

- 16 - 163 i

= - 16 - 163 i

(a - bi)^{2}

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