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

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

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simplify (i+sqrt(3))^4

simplify

(i + 3)^{4}

- 8 + 83 i

Solution steps

(i + 3)^{4}

Apply binomial theorem:

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

a = i, b = 3

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

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

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

Simplify

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

= 83 i - 8

Rewrite in standard complex form:

- 8 + 83 i

= - 8 + 83 i

(5 + 6) (3 + 5)

\frac{d}{d x} ((t) (a (x - h)^{2} + k))

- \frac{48 s ^{3} - 114 s ^{2} + s + 1}{6 s ^{2} ( s ^{2} - 2 s + 5 )}

\frac{6}{( 2 n - 1 ) ( 2 n + 1 )}

- 3 (x + Δ x)^{2}