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 - \frac{( x ^{2} )}{2}) \cdot (x - \frac{( x ^{3} )}{6})

l im x \Rightarrow 0 - (cos (2 x) + x^{2}) cot (3 x)

\frac{5 x ^{2} + 9 x + 3}{x ( x + 1 ) ^{2}}

\frac{8 x ^{\frac{2}{3}}}{( 4 x - 3 ) ^{\frac{2}{3}} ( 4 x + 3 ) ^{\frac{2}{3}}}

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