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

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

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

simplify

(3 i + 1)^{4}

- 8 - 83 i

Solution steps

(3 i + 1)^{4}

Apply binomial theorem:

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

a = 3 i, b = 1

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

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

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

Simplify

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

= - 83 i - 8

Rewrite in standard complex form:

- 8 - 83 i

= - 8 - 83 i

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