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

The solution to (3+sqrt(3)i)^4 is -72+72sqrt(3)i

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

simplify

(3 + 3 i)^{4}

- 72 + 723 i

Solution steps

(3 + 3 i)^{4}

Apply binomial theorem:

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

a = 3, b = 3 i

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

Expand summation

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

Simplify

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

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 3^{3} (3 i)^{1} : 1083 i

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 3^{2} (3 i)^{2} : 162 i^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 3^{1} (3 i)^{3} : 363 i^{3}

Simplify

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

= 81 + 1083 i + 162 i^{2} + 363 i^{3} + 9 i^{4}

Simplify

81 + 1083 i + 162 i^{2} + 363 i^{3} + 9 i^{4} : 723 i - 72

= 723 i - 72

Rewrite in standard complex form:

- 72 + 723 i

= - 72 + 723 i

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cos (2 (π - θ))