What is (1+isqrt(3))^8 ?

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

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simplify (1+isqrt(3))^8

simplify

(1 + i 3)^{8}

- 128 + 1283 i

Solution steps

(1 + i 3)^{8}

Apply binomial theorem:

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

a = 1, b = i 3

= i = 0 \sum 8 (i 8) \cdot 1^{(8 - i)} (i 3)^{i}

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

3 (2 + 5)

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