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

The solution to (sqrt(2)+sqrt(2)i)^5 is -16sqrt(2)-16sqrt(2)i

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

simplify

(2 + 2 i)^{5}

- 162 - 162 i

Solution steps

(2 + 2 i)^{5}

Apply binomial theorem:

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

a = 2, b = 2 i

= i = 0 \sum 5 (i 5) (2)^{(5 - i)} (2 i)^{i}

Expand summation

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

Simplify

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

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} (2)^{4} (2 i)^{1} : 202 i

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} (2)^{3} (2 i)^{2} : 402 i^{2}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} (2)^{2} (2 i)^{3} : 402 i^{3}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} (2)^{1} (2 i)^{4} : 202 i^{4}

Simplify

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

= 42 + 202 i + 402 i^{2} + 402 i^{3} + 202 i^{4} + 42 i^{5}

Simplify

42 + 202 i + 402 i^{2} + 402 i^{3} + 202 i^{4} + 42 i^{5} : - 162 - 162 i

= - 162 - 162 i

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(a - b) \frac{a + b}{a - b} - (a + b) \frac{a - b}{a + b} + (2 a - 2 b) \frac{1}{a - b}

(n - 1)!

(b c + d e) (b c - d e)

What is (sqrt(2)+sqrt(2)i)^5 ?
The solution to (sqrt(2)+sqrt(2)i)^5 is -16sqrt(2)-16sqrt(2)i