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Popular >

simplify (4-i)^5

Solution

simplify

(4 - i)^{5}

Solution

404 - 1121 i

Solution steps

(4 - i)^{5}

Apply binomial theorem:

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

a = 4, b = - i

= i = 0 \sum 5 (i 5) \cdot 4^{(5 - i)} (- i)^{i}

Expand summation

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

Simplify

\frac{5 !}{0 ! ( 5 - 0 ) !} \cdot 4^{5} (- i)^{0} : 1024

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} \cdot 4^{4} (- i)^{1} : - 1280 i

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} \cdot 4^{3} (- i)^{2} : 640 i^{2}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} \cdot 4^{2} (- i)^{3} : - 160 i^{3}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} \cdot 4^{1} (- i)^{4} : 20 i^{4}

Simplify

\frac{5 !}{5 ! ( 5 - 5 ) !} \cdot 4^{0} (- i)^{5} : - i^{5}

= 1024 - 1280 i + 640 i^{2} - 160 i^{3} + 20 i^{4} - i^{5}

Simplify

1024 - 1280 i + 640 i^{2} - 160 i^{3} + 20 i^{4} - i^{5} : - 1121 i + 404

= - 1121 i + 404

Rewrite in standard complex form:

404 - 1121 i

= 404 - 1121 i

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Frequently Asked Questions (FAQ)

What is (4-i)^5 ?
The solution to (4-i)^5 is 404-1121i