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

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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