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Popular >

simplify (1+2i)^7

Solution

simplify

(1 + 2 i)^{7}

Solution

29 + 278 i

Solution steps

(1 + 2 i)^{7}

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 7 (i 7) \cdot 1^{(7 - i)} (2 i)^{i}

Expand summation

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

Simplify

\frac{7 !}{0 ! ( 7 - 0 ) !} \cdot 1^{7} (2 i)^{0} : 1

Simplify

\frac{7 !}{1 ! ( 7 - 1 ) !} \cdot 1^{6} (2 i)^{1} : 14 i

Simplify

\frac{7 !}{2 ! ( 7 - 2 ) !} \cdot 1^{5} (2 i)^{2} : 84 i^{2}

Simplify

\frac{7 !}{3 ! ( 7 - 3 ) !} \cdot 1^{4} (2 i)^{3} : 280 i^{3}

Simplify

\frac{7 !}{4 ! ( 7 - 4 ) !} \cdot 1^{3} (2 i)^{4} : 560 i^{4}

Simplify

\frac{7 !}{5 ! ( 7 - 5 ) !} \cdot 1^{2} (2 i)^{5} : 672 i^{5}

Simplify

\frac{7 !}{6 ! ( 7 - 6 ) !} \cdot 1^{1} (2 i)^{6} : 448 i^{6}

Simplify

\frac{7 !}{7 ! ( 7 - 7 ) !} \cdot 1^{0} (2 i)^{7} : 128 i^{7}

= 1 + 14 i + 84 i^{2} + 280 i^{3} + 560 i^{4} + 672 i^{5} + 448 i^{6} + 128 i^{7}

Simplify

1 + 14 i + 84 i^{2} + 280 i^{3} + 560 i^{4} + 672 i^{5} + 448 i^{6} + 128 i^{7} : 278 i + 29

= 278 i + 29

Rewrite in standard complex form:

29 + 278 i

= 29 + 278 i

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Frequently Asked Questions (FAQ)

What is (1+2i)^7 ?
The solution to (1+2i)^7 is 29+278i