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Popular >

simplify (-2-2i)^5

Solution

simplify

(- 2 - 2 i)^{5}

Solution

128 + 128 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} : - 32

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} (- 2)^{4} (- 2 i)^{1} : - 160 i

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} (- 2)^{3} (- 2 i)^{2} : - 320 i^{2}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} (- 2)^{2} (- 2 i)^{3} : - 320 i^{3}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} (- 2)^{1} (- 2 i)^{4} : - 160 i^{4}

Simplify

\frac{5 !}{5 ! ( 5 - 5 ) !} (- 2)^{0} (- 2 i)^{5} : - 32 i^{5}

= - 32 - 160 i - 320 i^{2} - 320 i^{3} - 160 i^{4} - 32 i^{5}

Simplify

- 32 - 160 i - 320 i^{2} - 320 i^{3} - 160 i^{4} - 32 i^{5} : 128 i + 128

= 128 i + 128

Rewrite in standard complex form:

128 + 128 i

= 128 + 128 i

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Frequently Asked Questions (FAQ)

What is (-2-2i)^5 ?
The solution to (-2-2i)^5 is 128+128i