What is (2-2isqrt(3))^4 ?

The solution to (2-2isqrt(3))^4 is -128+128sqrt(3)i

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simplify (2-2isqrt(3))^4

simplify

(2 - 2 i 3)^{4}

- 128 + 1283 i

Solution steps

(2 - 2 i 3)^{4}

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 4 (i 4) \cdot 2^{(4 - i)} (- 2 i 3)^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 2^{4} (- 2 i 3)^{0} : 16

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 2^{3} (- 2 i 3)^{1} : - 643 i

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 2^{2} (- 2 i 3)^{2} : 288 i^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 2^{1} (- 2 i 3)^{3} : - 1923 i^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 2^{0} (- 2 i 3)^{4} : 144 i^{4}

= 16 - 643 i + 288 i^{2} - 1923 i^{3} + 144 i^{4}

Simplify

16 - 643 i + 288 i^{2} - 1923 i^{3} + 144 i^{4} : 1283 i - 128

= 1283 i - 128

Rewrite in standard complex form:

- 128 + 1283 i

= - 128 + 1283 i

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