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Popular >

simplify (1-2i)^8

Solution

simplify

(1 - 2 i)^{8}

Solution

- 527 - 336 i

Solution steps

(1 - 2 i)^{8}

Apply binomial theorem:

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

a = 1, b = - 2 i

= i = 0 \sum 8 (i 8) \cdot 1^{(8 - i)} (- 2 i)^{i}

Expand summation

= \frac{8 !}{0 ! ( 8 - 0 ) !} \cdot 1^{8} (- 2 i)^{0} + \frac{8 !}{1 ! ( 8 - 1 ) !} \cdot 1^{7} (- 2 i)^{1} + \frac{8 !}{2 ! ( 8 - 2 ) !} \cdot 1^{6} (- 2 i)^{2} + \frac{8 !}{3 ! ( 8 - 3 ) !} \cdot 1^{5} (- 2 i)^{3} + \frac{8 !}{4 ! ( 8 - 4 ) !} \cdot 1^{4} (- 2 i)^{4} + \frac{8 !}{5 ! ( 8 - 5 ) !} \cdot 1^{3} (- 2 i)^{5} + \frac{8 !}{6 ! ( 8 - 6 ) !} \cdot 1^{2} (- 2 i)^{6} + \frac{8 !}{7 ! ( 8 - 7 ) !} \cdot 1^{1} (- 2 i)^{7} + \frac{8 !}{8 ! ( 8 - 8 ) !} \cdot 1^{0} (- 2 i)^{8}

Simplify

\frac{8 !}{0 ! ( 8 - 0 ) !} \cdot 1^{8} (- 2 i)^{0} : 1

Simplify

\frac{8 !}{1 ! ( 8 - 1 ) !} \cdot 1^{7} (- 2 i)^{1} : - 16 i

Simplify

\frac{8 !}{2 ! ( 8 - 2 ) !} \cdot 1^{6} (- 2 i)^{2} : 112 i^{2}

Simplify

\frac{8 !}{3 ! ( 8 - 3 ) !} \cdot 1^{5} (- 2 i)^{3} : - 448 i^{3}

Simplify

\frac{8 !}{4 ! ( 8 - 4 ) !} \cdot 1^{4} (- 2 i)^{4} : 1120 i^{4}

Simplify

\frac{8 !}{5 ! ( 8 - 5 ) !} \cdot 1^{3} (- 2 i)^{5} : - 1792 i^{5}

Simplify

\frac{8 !}{6 ! ( 8 - 6 ) !} \cdot 1^{2} (- 2 i)^{6} : 1792 i^{6}

Simplify

\frac{8 !}{7 ! ( 8 - 7 ) !} \cdot 1^{1} (- 2 i)^{7} : - 1024 i^{7}

Simplify

\frac{8 !}{8 ! ( 8 - 8 ) !} \cdot 1^{0} (- 2 i)^{8} : 256 i^{8}

= 1 - 16 i + 112 i^{2} - 448 i^{3} + 1120 i^{4} - 1792 i^{5} + 1792 i^{6} - 1024 i^{7} + 256 i^{8}

Simplify

1 - 16 i + 112 i^{2} - 448 i^{3} + 1120 i^{4} - 1792 i^{5} + 1792 i^{6} - 1024 i^{7} + 256 i^{8} : - 336 i - 527

= - 336 i - 527

Rewrite in standard complex form:

- 527 - 336 i

= - 527 - 336 i

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Frequently Asked Questions (FAQ)

What is (1-2i)^8 ?
The solution to (1-2i)^8 is -527-336i