What is (1+sqrt(3)i)^8 ?

The solution to (1+sqrt(3)i)^8 is -128+128sqrt(3)i

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simplify (1+sqrt(3)i)^8

simplify

(1 + 3 i)^{8}

- 128 + 1283 i

Solution steps

(1 + 3 i)^{8}

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 8 (i 8) \cdot 1^{(8 - i)} (3 i)^{i}

Expand summation

= \frac{8 !}{0 ! ( 8 - 0 ) !} \cdot 1^{8} (3 i)^{0} + \frac{8 !}{1 ! ( 8 - 1 ) !} \cdot 1^{7} (3 i)^{1} + \frac{8 !}{2 ! ( 8 - 2 ) !} \cdot 1^{6} (3 i)^{2} + \frac{8 !}{3 ! ( 8 - 3 ) !} \cdot 1^{5} (3 i)^{3} + \frac{8 !}{4 ! ( 8 - 4 ) !} \cdot 1^{4} (3 i)^{4} + \frac{8 !}{5 ! ( 8 - 5 ) !} \cdot 1^{3} (3 i)^{5} + \frac{8 !}{6 ! ( 8 - 6 ) !} \cdot 1^{2} (3 i)^{6} + \frac{8 !}{7 ! ( 8 - 7 ) !} \cdot 1^{1} (3 i)^{7} + \frac{8 !}{8 ! ( 8 - 8 ) !} \cdot 1^{0} (3 i)^{8}

Simplify

\frac{8 !}{0 ! ( 8 - 0 ) !} \cdot 1^{8} (3 i)^{0} : 1

Simplify

\frac{8 !}{1 ! ( 8 - 1 ) !} \cdot 1^{7} (3 i)^{1} : 83 i

Simplify

\frac{8 !}{2 ! ( 8 - 2 ) !} \cdot 1^{6} (3 i)^{2} : 84 i^{2}

Simplify

\frac{8 !}{3 ! ( 8 - 3 ) !} \cdot 1^{5} (3 i)^{3} : 1683 i^{3}

Simplify

\frac{8 !}{4 ! ( 8 - 4 ) !} \cdot 1^{4} (3 i)^{4} : 630 i^{4}

Simplify

\frac{8 !}{5 ! ( 8 - 5 ) !} \cdot 1^{3} (3 i)^{5} : 5043 i^{5}

Simplify

\frac{8 !}{6 ! ( 8 - 6 ) !} \cdot 1^{2} (3 i)^{6} : 756 i^{6}

Simplify

\frac{8 !}{7 ! ( 8 - 7 ) !} \cdot 1^{1} (3 i)^{7} : 2163 i^{7}

Simplify

\frac{8 !}{8 ! ( 8 - 8 ) !} \cdot 1^{0} (3 i)^{8} : 81 i^{8}

= 1 + 83 i + 84 i^{2} + 1683 i^{3} + 630 i^{4} + 5043 i^{5} + 756 i^{6} + 2163 i^{7} + 81 i^{8}

Simplify

1 + 83 i + 84 i^{2} + 1683 i^{3} + 630 i^{4} + 5043 i^{5} + 756 i^{6} + 2163 i^{7} + 81 i^{8} : 1283 i - 128

= 1283 i - 128

Rewrite in standard complex form:

- 128 + 1283 i

= - 128 + 1283 i

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