What is (1+sqrt(3)i)^4 ?

The solution to (1+sqrt(3)i)^4 is -8-8sqrt(3)i

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simplify (1+sqrt(3)i)^4

simplify

(1 + 3 i)^{4}

- 8 - 83 i

Solution steps

(1 + 3 i)^{4}

Apply binomial theorem:

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

a = 1, b = 3 i

= i = 0 \sum 4 (i 4) \cdot 1^{(4 - i)} (3 i)^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 1^{4} (3 i)^{0} : 1

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 1^{3} (3 i)^{1} : 43 i

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 1^{2} (3 i)^{2} : 18 i^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 1^{1} (3 i)^{3} : 123 i^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 1^{0} (3 i)^{4} : 9 i^{4}

= 1 + 43 i + 18 i^{2} + 123 i^{3} + 9 i^{4}

Simplify

1 + 43 i + 18 i^{2} + 123 i^{3} + 9 i^{4} : - 83 i - 8

= - 83 i - 8

Rewrite in standard complex form:

- 8 - 83 i

= - 8 - 83 i

(cos (x) - x sin (x))^{2}

(x + y) (x + y)

\frac{2 x - 12}{( x - 4 ) ( x - 5 )}

(a - x)^{4}

\frac{x}{( x + 1 ) ^{2}}