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Popular >

simplify (4-2i)^5

Solution

simplify

(4 - 2 i)^{5}

Solution

- 1216 - 1312 i

Solution steps

(4 - 2 i)^{5}

Apply binomial theorem:

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

a = 4, b = - 2 i

= i = 0 \sum 5 (i 5) \cdot 4^{(5 - i)} (- 2 i)^{i}

Expand summation

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

Simplify

\frac{5 !}{0 ! ( 5 - 0 ) !} \cdot 4^{5} (- 2 i)^{0} : 1024

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} \cdot 4^{4} (- 2 i)^{1} : - 2560 i

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} \cdot 4^{3} (- 2 i)^{2} : 2560 i^{2}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} \cdot 4^{2} (- 2 i)^{3} : - 1280 i^{3}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} \cdot 4^{1} (- 2 i)^{4} : 320 i^{4}

Simplify

\frac{5 !}{5 ! ( 5 - 5 ) !} \cdot 4^{0} (- 2 i)^{5} : - 32 i^{5}

= 1024 - 2560 i + 2560 i^{2} - 1280 i^{3} + 320 i^{4} - 32 i^{5}

Simplify

1024 - 2560 i + 2560 i^{2} - 1280 i^{3} + 320 i^{4} - 32 i^{5} : - 1312 i - 1216

= - 1312 i - 1216

Rewrite in standard complex form:

- 1216 - 1312 i

= - 1216 - 1312 i

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Frequently Asked Questions (FAQ)

What is (4-2i)^5 ?
The solution to (4-2i)^5 is -1216-1312i