What is (-3+3isqrt(3))^4 ?

The solution to (-3+3isqrt(3))^4 is -648+648sqrt(3)i

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simplify (-3+3isqrt(3))^4

simplify

(- 3 + 3 i 3)^{4}

- 648 + 6483 i

Solution steps

(- 3 + 3 i 3)^{4}

Apply binomial theorem:

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

a = - 3, b = 3 i 3

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

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (- 3)^{4} (3 i 3)^{0} : 81

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (- 3)^{3} (3 i 3)^{1} : - 3243 i

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (- 3)^{2} (3 i 3)^{2} : 1458 i^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (- 3)^{1} (3 i 3)^{3} : - 9723 i^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (- 3)^{0} (3 i 3)^{4} : 729 i^{4}

= 81 - 3243 i + 1458 i^{2} - 9723 i^{3} + 729 i^{4}

Simplify

81 - 3243 i + 1458 i^{2} - 9723 i^{3} + 729 i^{4} : 6483 i - 648

= 6483 i - 648

Rewrite in standard complex form:

- 648 + 6483 i

= - 648 + 6483 i

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