What is (2+sqrt(5)i)^7 ?

The solution to (2+sqrt(5)i)^7 is 2018-377sqrt(5)i

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simplify (2+sqrt(5)i)^7

simplify

(2 + 5 i)^{7}

2018 - 3775 i

Solution steps

(2 + 5 i)^{7}

Apply binomial theorem:

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

a = 2, b = 5 i

= i = 0 \sum 7 (i 7) \cdot 2^{(7 - i)} (5 i)^{i}

Expand summation

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

Simplify

\frac{7 !}{0 ! ( 7 - 0 ) !} \cdot 2^{7} (5 i)^{0} : 128

Simplify

\frac{7 !}{1 ! ( 7 - 1 ) !} \cdot 2^{6} (5 i)^{1} : 4485 i

Simplify

\frac{7 !}{2 ! ( 7 - 2 ) !} \cdot 2^{5} (5 i)^{2} : 3360 i^{2}

Simplify

\frac{7 !}{3 ! ( 7 - 3 ) !} \cdot 2^{4} (5 i)^{3} : 28005 i^{3}

Simplify

\frac{7 !}{4 ! ( 7 - 4 ) !} \cdot 2^{3} (5 i)^{4} : 7000 i^{4}

Simplify

\frac{7 !}{5 ! ( 7 - 5 ) !} \cdot 2^{2} (5 i)^{5} : 21005 i^{5}

Simplify

\frac{7 !}{6 ! ( 7 - 6 ) !} \cdot 2^{1} (5 i)^{6} : 1750 i^{6}

Simplify

\frac{7 !}{7 ! ( 7 - 7 ) !} \cdot 2^{0} (5 i)^{7} : 1255 i^{7}

= 128 + 4485 i + 3360 i^{2} + 28005 i^{3} + 7000 i^{4} + 21005 i^{5} + 1750 i^{6} + 1255 i^{7}

Simplify

128 + 4485 i + 3360 i^{2} + 28005 i^{3} + 7000 i^{4} + 21005 i^{5} + 1750 i^{6} + 1255 i^{7} : - 3775 i + 2018

= - 3775 i + 2018

Rewrite in standard complex form:

2018 - 3775 i

= 2018 - 3775 i

- (\frac{2 \cdot 3 ^{2}}{3 \cdot 5}) + 8 \cdot 3 - 4 + 4 \cdot 11

b - b

(2 x - 5) (3 x^{2} - 4 x - 8)

\frac{121}{4}

\frac{x ^{\frac{1}{2}} y ^{\frac{2}{3}}}{x ^{\frac{1}{4}} y ^{\frac{1}{6}}}