What is (sqrt(3)+i)^{10} ?

The solution to (sqrt(3)+i)^{10} is 512-512sqrt(3)i

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simplify (sqrt(3)+i)^{10}

simplify

(3 + i)^{10}

512 - 5123 i

Solution steps

(3 + i)^{10}

Apply binomial theorem:

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

a = 3, b = i

= i = 0 \sum 10 (i 10) (3)^{(10 - i)} i^{i}

Expand summation

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

Simplify

\frac{10 !}{0 ! ( 10 - 0 ) !} (3)^{10} i^{0} : 243

Simplify

\frac{10 !}{1 ! ( 10 - 1 ) !} (3)^{9} i^{1} : 8103 i

Simplify

\frac{10 !}{2 ! ( 10 - 2 ) !} (3)^{8} i^{2} : 3645 i^{2}

Simplify

\frac{10 !}{3 ! ( 10 - 3 ) !} (3)^{7} i^{3} : 32403 i^{3}

Simplify

\frac{10 !}{4 ! ( 10 - 4 ) !} (3)^{6} i^{4} : 5670 i^{4}

Simplify

\frac{10 !}{5 ! ( 10 - 5 ) !} (3)^{5} i^{5} : 22683 i^{5}

Simplify

\frac{10 !}{6 ! ( 10 - 6 ) !} (3)^{4} i^{6} : 1890 i^{6}

Simplify

\frac{10 !}{7 ! ( 10 - 7 ) !} (3)^{3} i^{7} : 3603 i^{7}

Simplify

\frac{10 !}{8 ! ( 10 - 8 ) !} (3)^{2} i^{8} : 135 i^{8}

Simplify

\frac{10 !}{9 ! ( 10 - 9 ) !} (3)^{1} i^{9} : 103 i^{9}

Simplify

\frac{10 !}{10 ! ( 10 - 10 ) !} (3)^{0} i^{10} : i^{10}

= 243 + 8103 i + 3645 i^{2} + 32403 i^{3} + 5670 i^{4} + 22683 i^{5} + 1890 i^{6} + 3603 i^{7} + 135 i^{8} + 103 i^{9} + i^{10}

Simplify

243 + 8103 i + 3645 i^{2} + 32403 i^{3} + 5670 i^{4} + 22683 i^{5} + 1890 i^{6} + 3603 i^{7} + 135 i^{8} + 103 i^{9} + i^{10} : - 5123 i + 512

= - 5123 i + 512

Rewrite in standard complex form:

512 - 5123 i

= 512 - 5123 i

e^{x} (- e^{- x} x - e^{- x})

\frac{( 6 - 3 x ) ^{2}}{4}

\frac{s - 1}{( s + 2 ) ( s ^{2} + 4 )}

\frac{2 ( x - 1 ) ^{2}}{( x - 2 ) ( x + 1 )}

(1 + i x)^{2}