What is (2+2sqrt(3)i)^5 ?

The solution to (2+2sqrt(3)i)^5 is 512-512sqrt(3)i

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simplify (2+2sqrt(3)i)^5

simplify

(2 + 23 i)^{5}

512 - 5123 i

Solution steps

(2 + 23 i)^{5}

Apply binomial theorem:

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

a = 2, b = 23 i

= i = 0 \sum 5 (i 5) \cdot 2^{(5 - i)} (23 i)^{i}

Expand summation

= \frac{5 !}{0 ! ( 5 - 0 ) !} \cdot 2^{5} (23 i)^{0} + \frac{5 !}{1 ! ( 5 - 1 ) !} \cdot 2^{4} (23 i)^{1} + \frac{5 !}{2 ! ( 5 - 2 ) !} \cdot 2^{3} (23 i)^{2} + \frac{5 !}{3 ! ( 5 - 3 ) !} \cdot 2^{2} (23 i)^{3} + \frac{5 !}{4 ! ( 5 - 4 ) !} \cdot 2^{1} (23 i)^{4} + \frac{5 !}{5 ! ( 5 - 5 ) !} \cdot 2^{0} (23 i)^{5}

Simplify

\frac{5 !}{0 ! ( 5 - 0 ) !} \cdot 2^{5} (23 i)^{0} : 32

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} \cdot 2^{4} (23 i)^{1} : 1603 i

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} \cdot 2^{3} (23 i)^{2} : 960 i^{2}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} \cdot 2^{2} (23 i)^{3} : 9603 i^{3}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} \cdot 2^{1} (23 i)^{4} : 1440 i^{4}

Simplify

\frac{5 !}{5 ! ( 5 - 5 ) !} \cdot 2^{0} (23 i)^{5} : 2883 i^{5}

= 32 + 1603 i + 960 i^{2} + 9603 i^{3} + 1440 i^{4} + 2883 i^{5}

Simplify

32 + 1603 i + 960 i^{2} + 9603 i^{3} + 1440 i^{4} + 2883 i^{5} : - 5123 i + 512

= - 5123 i + 512

Rewrite in standard complex form:

512 - 5123 i

= 512 - 5123 i

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