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Popular >

simplify (1+i)^5

Solution

simplify

(1 + i)^{5}

Solution

- 4 - 4 i

Solution steps

(1 + i)^{5}

Apply binomial theorem:

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

a = 1, b = i

= i = 0 \sum 5 (i 5) \cdot 1^{(5 - i)} i^{i}

Expand summation

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

Simplify

\frac{5 !}{0 ! ( 5 - 0 ) !} \cdot 1^{5} i^{0} : 1

Simplify

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

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} \cdot 1^{3} i^{2} : 10 i^{2}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} \cdot 1^{2} i^{3} : 10 i^{3}

Simplify

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

Simplify

\frac{5 !}{5 ! ( 5 - 5 ) !} \cdot 1^{0} i^{5} : i^{5}

= 1 + 5 i + 10 i^{2} + 10 i^{3} + 5 i^{4} + i^{5}

Simplify

1 + 5 i + 10 i^{2} + 10 i^{3} + 5 i^{4} + i^{5} : - 4 i - 4

= - 4 i - 4

Rewrite in standard complex form:

- 4 - 4 i

= - 4 - 4 i

Popular Examples

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(x + i y)^{3}

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(e^{x} + e^{- x})^{2}

expand (1-cos(x))^2 expand

(1 - cos (x))^{2}

Frequently Asked Questions (FAQ)

What is (1+i)^5 ?
The solution to (1+i)^5 is -4-4i