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Popular >

simplify (1-i)^9

Solution

simplify

(1 - i)^{9}

Solution

16 - 16 i

Solution steps

(1 - i)^{9}

Apply binomial theorem:

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

a = 1, b = - i

= i = 0 \sum 9 (i 9) \cdot 1^{(9 - i)} (- i)^{i}

Expand summation

= \frac{9 !}{0 ! ( 9 - 0 ) !} \cdot 1^{9} (- i)^{0} + \frac{9 !}{1 ! ( 9 - 1 ) !} \cdot 1^{8} (- i)^{1} + \frac{9 !}{2 ! ( 9 - 2 ) !} \cdot 1^{7} (- i)^{2} + \frac{9 !}{3 ! ( 9 - 3 ) !} \cdot 1^{6} (- i)^{3} + \frac{9 !}{4 ! ( 9 - 4 ) !} \cdot 1^{5} (- i)^{4} + \frac{9 !}{5 ! ( 9 - 5 ) !} \cdot 1^{4} (- i)^{5} + \frac{9 !}{6 ! ( 9 - 6 ) !} \cdot 1^{3} (- i)^{6} + \frac{9 !}{7 ! ( 9 - 7 ) !} \cdot 1^{2} (- i)^{7} + \frac{9 !}{8 ! ( 9 - 8 ) !} \cdot 1^{1} (- i)^{8} + \frac{9 !}{9 ! ( 9 - 9 ) !} \cdot 1^{0} (- i)^{9}

Simplify

\frac{9 !}{0 ! ( 9 - 0 ) !} \cdot 1^{9} (- i)^{0} : 1

Simplify

\frac{9 !}{1 ! ( 9 - 1 ) !} \cdot 1^{8} (- i)^{1} : - 9 i

Simplify

\frac{9 !}{2 ! ( 9 - 2 ) !} \cdot 1^{7} (- i)^{2} : 36 i^{2}

Simplify

\frac{9 !}{3 ! ( 9 - 3 ) !} \cdot 1^{6} (- i)^{3} : - 84 i^{3}

Simplify

\frac{9 !}{4 ! ( 9 - 4 ) !} \cdot 1^{5} (- i)^{4} : 126 i^{4}

Simplify

\frac{9 !}{5 ! ( 9 - 5 ) !} \cdot 1^{4} (- i)^{5} : - 126 i^{5}

Simplify

\frac{9 !}{6 ! ( 9 - 6 ) !} \cdot 1^{3} (- i)^{6} : 84 i^{6}

Simplify

\frac{9 !}{7 ! ( 9 - 7 ) !} \cdot 1^{2} (- i)^{7} : - 36 i^{7}

Simplify

\frac{9 !}{8 ! ( 9 - 8 ) !} \cdot 1^{1} (- i)^{8} : 9 i^{8}

Simplify

\frac{9 !}{9 ! ( 9 - 9 ) !} \cdot 1^{0} (- i)^{9} : - i^{9}

= 1 - 9 i + 36 i^{2} - 84 i^{3} + 126 i^{4} - 126 i^{5} + 84 i^{6} - 36 i^{7} + 9 i^{8} - i^{9}

Simplify

1 - 9 i + 36 i^{2} - 84 i^{3} + 126 i^{4} - 126 i^{5} + 84 i^{6} - 36 i^{7} + 9 i^{8} - i^{9} : - 16 i + 16

= - 16 i + 16

Rewrite in standard complex form:

16 - 16 i

= 16 - 16 i

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Frequently Asked Questions (FAQ)

What is (1-i)^9 ?
The solution to (1-i)^9 is 16-16i