What is (sqrt(2)+sqrt(6)i)^4 ?

The solution to (sqrt(2)+sqrt(6)i)^4 is -32-32sqrt(3)i

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simplify (sqrt(2)+sqrt(6)i)^4

simplify

(2 + 6 i)^{4}

- 32 - 323 i

Solution steps

(2 + 6 i)^{4}

Apply binomial theorem:

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

a = 2, b = 6 i

= i = 0 \sum 4 (i 4) (2)^{(4 - i)} (6 i)^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} (2)^{4} (6 i)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (2)^{3} (6 i)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (2)^{2} (6 i)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (2)^{1} (6 i)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (2)^{0} (6 i)^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (2)^{4} (6 i)^{0} : 4

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (2)^{3} (6 i)^{1} : 163 i

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (2)^{2} (6 i)^{2} : 72 i^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (2)^{1} (6 i)^{3} : 483 i^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (2)^{0} (6 i)^{4} : 36 i^{4}

= 4 + 163 i + 72 i^{2} + 483 i^{3} + 36 i^{4}

Simplify

4 + 163 i + 72 i^{2} + 483 i^{3} + 36 i^{4} : - 323 i - 32

= - 323 i - 32

Rewrite in standard complex form:

- 32 - 323 i

= - 32 - 323 i

1 - 2 (x - \frac{π}{4})

\frac{d}{d x} ((x - 2) (x - 5)^{3}) + 12

x (1 - \frac{x}{2})

4 (3 + 3)

What is (sqrt(2)+sqrt(6)i)^4 ?
The solution to (sqrt(2)+sqrt(6)i)^4 is -32-32sqrt(3)i