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Popular >

simplify (i+2)^5

Solution

simplify

(i + 2)^{5}

Solution

- 38 + 41 i

Solution steps

(i + 2)^{5}

Apply binomial theorem:

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

a = i, b = 2

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

Expand summation

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

Simplify

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

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} i^{4} \cdot 2^{1} : 10 i^{4}

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} i^{3} \cdot 2^{2} : 40 i^{3}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} i^{2} \cdot 2^{3} : 80 i^{2}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} i^{1} \cdot 2^{4} : 80 i

Simplify

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

= i^{5} + 10 i^{4} + 40 i^{3} + 80 i^{2} + 80 i + 32

Simplify

i^{5} + 10 i^{4} + 40 i^{3} + 80 i^{2} + 80 i + 32 : 41 i - 38

= 41 i - 38

Rewrite in standard complex form:

- 38 + 41 i

= - 38 + 41 i

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Frequently Asked Questions (FAQ)

What is (i+2)^5 ?
The solution to (i+2)^5 is -38+41i