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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) \cdot 2^{(5 - i)} (2 i)^{i}

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

\frac{5 !}{5 ! ( 5 - 5 ) !} \cdot 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