What is (2-2isqrt(3))^5 ?

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

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simplify (2-2isqrt(3))^5

simplify

(2 - 2 i 3)^{5}

512 + 5123 i

Solution steps

(2 - 2 i 3)^{5}

Apply binomial theorem:

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

a = 2, b = - 2 i 3

= i = 0 \sum 5 (i 5) \cdot 2^{(5 - i)} (- 2 i 3)^{i}

Expand summation

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

Simplify

\frac{5 !}{0 ! ( 5 - 0 ) !} \cdot 2^{5} (- 2 i 3)^{0} : 32

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} \cdot 2^{4} (- 2 i 3)^{1} : - 1603 i

Simplify

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

Simplify

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

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} \cdot 2^{1} (- 2 i 3)^{4} : 1440 i^{4}

Simplify

\frac{5 !}{5 ! ( 5 - 5 ) !} \cdot 2^{0} (- 2 i 3)^{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

\frac{( s + 1 ) ^{3}}{s ^{5}}

(1 - x) e^{x}

ln (a + 1) \cdot (e^{a} + a \cdot e^{a})

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

\frac{5}{( x - 1 ) ( x + 4 )}