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

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

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

simplify

(33 - 3 i)^{4}

- 648 - 6483 i

Solution steps

(33 - 3 i)^{4}

Apply binomial theorem:

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

a = 33, b = - 3 i

= i = 0 \sum 4 (i 4) (33)^{(4 - i)} (- 3 i)^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} (33)^{4} (- 3 i)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (33)^{3} (- 3 i)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (33)^{2} (- 3 i)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (33)^{1} (- 3 i)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (33)^{0} (- 3 i)^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (33)^{4} (- 3 i)^{0} : 729

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (33)^{3} (- 3 i)^{1} : - 9723 i

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (33)^{2} (- 3 i)^{2} : 1458 i^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (33)^{1} (- 3 i)^{3} : - 3243 i^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (33)^{0} (- 3 i)^{4} : 81 i^{4}

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

Simplify

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

= - 6483 i - 648

Rewrite in standard complex form:

- 648 - 6483 i

= - 648 - 6483 i

21 \cdot 8

(5 - 3 i + 5 i^{5}) - (- 3 + 8 i)

\frac{4}{7} + \frac{3}{5}

(cos (2 t) + 2 t sin (2 t))^{2}

\frac{2 - 2}{3 + 2}