What is (-1+sqrt(2)i)^9 ?

The solution to (-1+sqrt(2)i)^9 is 95+73sqrt(2)i

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simplify (-1+sqrt(2)i)^9

simplify

(- 1 + 2 i)^{9}

95 + 732 i

Solution steps

(- 1 + 2 i)^{9}

Apply binomial theorem:

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

a = - 1, b = 2 i

= i = 0 \sum 9 (i 9) (- 1)^{(9 - i)} (2 i)^{i}

Expand summation

= \frac{9 !}{0 ! ( 9 - 0 ) !} (- 1)^{9} (2 i)^{0} + \frac{9 !}{1 ! ( 9 - 1 ) !} (- 1)^{8} (2 i)^{1} + \frac{9 !}{2 ! ( 9 - 2 ) !} (- 1)^{7} (2 i)^{2} + \frac{9 !}{3 ! ( 9 - 3 ) !} (- 1)^{6} (2 i)^{3} + \frac{9 !}{4 ! ( 9 - 4 ) !} (- 1)^{5} (2 i)^{4} + \frac{9 !}{5 ! ( 9 - 5 ) !} (- 1)^{4} (2 i)^{5} + \frac{9 !}{6 ! ( 9 - 6 ) !} (- 1)^{3} (2 i)^{6} + \frac{9 !}{7 ! ( 9 - 7 ) !} (- 1)^{2} (2 i)^{7} + \frac{9 !}{8 ! ( 9 - 8 ) !} (- 1)^{1} (2 i)^{8} + \frac{9 !}{9 ! ( 9 - 9 ) !} (- 1)^{0} (2 i)^{9}

Simplify

\frac{9 !}{0 ! ( 9 - 0 ) !} (- 1)^{9} (2 i)^{0} : - 1

Simplify

\frac{9 !}{1 ! ( 9 - 1 ) !} (- 1)^{8} (2 i)^{1} : 92 i

Simplify

\frac{9 !}{2 ! ( 9 - 2 ) !} (- 1)^{7} (2 i)^{2} : - 72 i^{2}

Simplify

\frac{9 !}{3 ! ( 9 - 3 ) !} (- 1)^{6} (2 i)^{3} : 1682 i^{3}

Simplify

\frac{9 !}{4 ! ( 9 - 4 ) !} (- 1)^{5} (2 i)^{4} : - 504 i^{4}

Simplify

\frac{9 !}{5 ! ( 9 - 5 ) !} (- 1)^{4} (2 i)^{5} : 5042 i^{5}

Simplify

\frac{9 !}{6 ! ( 9 - 6 ) !} (- 1)^{3} (2 i)^{6} : - 672 i^{6}

Simplify

\frac{9 !}{7 ! ( 9 - 7 ) !} (- 1)^{2} (2 i)^{7} : 2882 i^{7}

Simplify

\frac{9 !}{8 ! ( 9 - 8 ) !} (- 1)^{1} (2 i)^{8} : - 144 i^{8}

Simplify

\frac{9 !}{9 ! ( 9 - 9 ) !} (- 1)^{0} (2 i)^{9} : 162 i^{9}

= - 1 + 92 i - 72 i^{2} + 1682 i^{3} - 504 i^{4} + 5042 i^{5} - 672 i^{6} + 2882 i^{7} - 144 i^{8} + 162 i^{9}

Simplify

- 1 + 92 i - 72 i^{2} + 1682 i^{3} - 504 i^{4} + 5042 i^{5} - 672 i^{6} + 2882 i^{7} - 144 i^{8} + 162 i^{9} : 732 i + 95

= 732 i + 95

Rewrite in standard complex form:

95 + 732 i

= 95 + 732 i

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