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Popular >

simplify (1+i)^{11}

Solution

simplify

(1 + i)^{11}

Solution

- 32 + 32 i

Solution steps

(1 + i)^{11}

Apply binomial theorem:

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

a = 1, b = i

= i = 0 \sum 11 (i 11) \cdot 1^{(11 - i)} i^{i}

Expand summation

= \frac{11 !}{0 ! ( 11 - 0 ) !} \cdot 1^{11} i^{0} + \frac{11 !}{1 ! ( 11 - 1 ) !} \cdot 1^{10} i^{1} + \frac{11 !}{2 ! ( 11 - 2 ) !} \cdot 1^{9} i^{2} + \frac{11 !}{3 ! ( 11 - 3 ) !} \cdot 1^{8} i^{3} + \frac{11 !}{4 ! ( 11 - 4 ) !} \cdot 1^{7} i^{4} + \frac{11 !}{5 ! ( 11 - 5 ) !} \cdot 1^{6} i^{5} + \frac{11 !}{6 ! ( 11 - 6 ) !} \cdot 1^{5} i^{6} + \frac{11 !}{7 ! ( 11 - 7 ) !} \cdot 1^{4} i^{7} + \frac{11 !}{8 ! ( 11 - 8 ) !} \cdot 1^{3} i^{8} + \frac{11 !}{9 ! ( 11 - 9 ) !} \cdot 1^{2} i^{9} + \frac{11 !}{10 ! ( 11 - 10 ) !} \cdot 1^{1} i^{10} + \frac{11 !}{11 ! ( 11 - 11 ) !} \cdot 1^{0} i^{11}

Simplify

\frac{11 !}{0 ! ( 11 - 0 ) !} \cdot 1^{11} i^{0} : 1

Simplify

\frac{11 !}{1 ! ( 11 - 1 ) !} \cdot 1^{10} i^{1} : 11 i

Simplify

\frac{11 !}{2 ! ( 11 - 2 ) !} \cdot 1^{9} i^{2} : 55 i^{2}

Simplify

\frac{11 !}{3 ! ( 11 - 3 ) !} \cdot 1^{8} i^{3} : 165 i^{3}

Simplify

\frac{11 !}{4 ! ( 11 - 4 ) !} \cdot 1^{7} i^{4} : 330 i^{4}

Simplify

\frac{11 !}{5 ! ( 11 - 5 ) !} \cdot 1^{6} i^{5} : 462 i^{5}

Simplify

\frac{11 !}{6 ! ( 11 - 6 ) !} \cdot 1^{5} i^{6} : 462 i^{6}

Simplify

\frac{11 !}{7 ! ( 11 - 7 ) !} \cdot 1^{4} i^{7} : 330 i^{7}

Simplify

\frac{11 !}{8 ! ( 11 - 8 ) !} \cdot 1^{3} i^{8} : 165 i^{8}

Simplify

\frac{11 !}{9 ! ( 11 - 9 ) !} \cdot 1^{2} i^{9} : 55 i^{9}

Simplify

\frac{11 !}{10 ! ( 11 - 10 ) !} \cdot 1^{1} i^{10} : 11 i^{10}

Simplify

\frac{11 !}{11 ! ( 11 - 11 ) !} \cdot 1^{0} i^{11} : i^{11}

= 1 + 11 i + 55 i^{2} + 165 i^{3} + 330 i^{4} + 462 i^{5} + 462 i^{6} + 330 i^{7} + 165 i^{8} + 55 i^{9} + 11 i^{10} + i^{11}

Simplify

1 + 11 i + 55 i^{2} + 165 i^{3} + 330 i^{4} + 462 i^{5} + 462 i^{6} + 330 i^{7} + 165 i^{8} + 55 i^{9} + 11 i^{10} + i^{11} : 32 i - 32

= 32 i - 32

Rewrite in standard complex form:

- 32 + 32 i

= - 32 + 32 i

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Frequently Asked Questions (FAQ)

What is (1+i)^{11} ?
The solution to (1+i)^{11} is -32+32i