What is (1+sqrt(3)i)^7 ?

The solution to (1+sqrt(3)i)^7 is 64+64sqrt(3)i

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simplify (1+sqrt(3)i)^7

simplify

(1 + 3 i)^{7}

64 + 643 i

Solution steps

(1 + 3 i)^{7}

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 7 (i 7) \cdot 1^{(7 - i)} (3 i)^{i}

Expand summation

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

Simplify

\frac{7 !}{0 ! ( 7 - 0 ) !} \cdot 1^{7} (3 i)^{0} : 1

Simplify

\frac{7 !}{1 ! ( 7 - 1 ) !} \cdot 1^{6} (3 i)^{1} : 73 i

Simplify

\frac{7 !}{2 ! ( 7 - 2 ) !} \cdot 1^{5} (3 i)^{2} : 63 i^{2}

Simplify

\frac{7 !}{3 ! ( 7 - 3 ) !} \cdot 1^{4} (3 i)^{3} : 1053 i^{3}

Simplify

\frac{7 !}{4 ! ( 7 - 4 ) !} \cdot 1^{3} (3 i)^{4} : 315 i^{4}

Simplify

\frac{7 !}{5 ! ( 7 - 5 ) !} \cdot 1^{2} (3 i)^{5} : 1893 i^{5}

Simplify

\frac{7 !}{6 ! ( 7 - 6 ) !} \cdot 1^{1} (3 i)^{6} : 189 i^{6}

Simplify

\frac{7 !}{7 ! ( 7 - 7 ) !} \cdot 1^{0} (3 i)^{7} : 273 i^{7}

= 1 + 73 i + 63 i^{2} + 1053 i^{3} + 315 i^{4} + 1893 i^{5} + 189 i^{6} + 273 i^{7}

Simplify

1 + 73 i + 63 i^{2} + 1053 i^{3} + 315 i^{4} + 1893 i^{5} + 189 i^{6} + 273 i^{7} : 643 i + 64

= 643 i + 64

Rewrite in standard complex form:

64 + 643 i

= 64 + 643 i

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