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) (- 1)^{(4 - i)} (i 3)^{i}

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (- 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

(x + i)^{4}

(8 + 7) (9 + 2)

(x + 5 x^{2})^{2}

(1 + 2 x - 2)^{2}

\frac{( x + 1 ) ( x + 3 )}{2}