What is (sqrt(3)+i)^5 ?

The solution to (sqrt(3)+i)^5 is -16sqrt(3)+16i

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simplify (sqrt(3)+i)^5

simplify

(3 + i)^{5}

- 163 + 16 i

Solution steps

(3 + i)^{5}

Apply binomial theorem:

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

a = 3, b = i

= i = 0 \sum 5 (i 5) (3)^{(5 - i)} i^{i}

Expand summation

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

Simplify

\frac{5 !}{0 ! ( 5 - 0 ) !} (3)^{5} i^{0} : 93

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} (3)^{4} i^{1} : 45 i

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} (3)^{3} i^{2} : 303 i^{2}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} (3)^{2} i^{3} : 30 i^{3}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} (3)^{1} i^{4} : 53 i^{4}

Simplify

\frac{5 !}{5 ! ( 5 - 5 ) !} (3)^{0} i^{5} : i^{5}

= 93 + 45 i + 303 i^{2} + 30 i^{3} + 53 i^{4} + i^{5}

Simplify

93 + 45 i + 303 i^{2} + 30 i^{3} + 53 i^{4} + i^{5} : 16 i - 163

= 16 i - 163

Rewrite in standard complex form:

- 163 + 16 i

= - 163 + 16 i

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