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

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

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simplify (i+sqrt(3))^5

simplify

(i + 3)^{5}

- 163 + 16 i

Solution steps

(i + 3)^{5}

Apply binomial theorem:

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

a = i, b = 3

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

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

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

Simplify

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

= 16 i - 163

Rewrite in standard complex form:

- 163 + 16 i

= - 163 + 16 i

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