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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

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Frequently Asked Questions (FAQ)

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