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

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

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

simplify

(3 - 33 i)^{4}

- 648 + 6483 i

Solution steps

(3 - 33 i)^{4}

Apply binomial theorem:

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

a = 3, b = - 33 i

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

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

\frac{( p - 100 ) ^{2}}{2}

(\frac{x}{2} - \frac{2}{x})^{6}

\frac{1}{2} + \frac{3}{44} (5 (0.6) - \frac{( 0.6 ) ^{3}}{3})

(x^{2} - 3) d x

(n + i)^{2}