What is (1+isqrt(3))^4 ?

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

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simplify (1+isqrt(3))^4

simplify

(1 + i 3)^{4}

- 8 - 83 i

Solution steps

(1 + i 3)^{4}

Apply binomial theorem:

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

a = 1, b = i 3

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

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

(2 n + 2)!

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

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