What is (-1+sqrt(2)i)^5 ?

The solution to (-1+sqrt(2)i)^5 is -1-11sqrt(2)i

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simplify (-1+sqrt(2)i)^5

simplify

(- 1 + 2 i)^{5}

- 1 - 112 i

Solution steps

(- 1 + 2 i)^{5}

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 5 (i 5) (- 1)^{(5 - i)} (2 i)^{i}

Expand summation

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

Simplify

\frac{5 !}{0 ! ( 5 - 0 ) !} (- 1)^{5} (2 i)^{0} : - 1

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} (- 1)^{4} (2 i)^{1} : 52 i

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} (- 1)^{3} (2 i)^{2} : - 20 i^{2}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} (- 1)^{2} (2 i)^{3} : 202 i^{3}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} (- 1)^{1} (2 i)^{4} : - 20 i^{4}

Simplify

\frac{5 !}{5 ! ( 5 - 5 ) !} (- 1)^{0} (2 i)^{5} : 42 i^{5}

= - 1 + 52 i - 20 i^{2} + 202 i^{3} - 20 i^{4} + 42 i^{5}

Simplify

- 1 + 52 i - 20 i^{2} + 202 i^{3} - 20 i^{4} + 42 i^{5} : - 112 i - 1

= - 112 i - 1

Rewrite in standard complex form:

- 1 - 112 i

= - 1 - 112 i

\frac{2 s ^{3} - 9 s ^{2} - 24 s - 12}{( s + 1 ) ^{2} ( s - 1 ) ( s - 5 )}

(2 + 32) \cdot (- 41)

x (ln (x) + c)

(2 x i + 2 y j - z k) \cdot (aii +, + c kk) \cdot (2 x i + 2 y j - z k)

7 (x^{\frac{1}{3}} + 9)^{5}