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Popular >

simplify (2+i)^5

Solution

simplify

(2 + i)^{5}

Solution

- 38 + 41 i

Solution steps

(2 + i)^{5}

Apply binomial theorem:

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

a = 2, b = i

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

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

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

Simplify

32 + 80 i + 80 i^{2} + 40 i^{3} + 10 i^{4} + i^{5} : 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 (2+i)^5 ?
The solution to (2+i)^5 is -38+41i