What is (2+isqrt(3))^4 ?

The solution to (2+isqrt(3))^4 is -47+8sqrt(3)i

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simplify (2+isqrt(3))^4

simplify

(2 + i 3)^{4}

- 47 + 83 i

Solution steps

(2 + i 3)^{4}

Apply binomial theorem:

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

a = 2, b = i 3

= i = 0 \sum 4 (i 4) \cdot 2^{(4 - i)} (i 3)^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 2^{4} (i 3)^{0} : 16

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 2^{3} (i 3)^{1} : 323 i

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 2^{2} (i 3)^{2} : 72 i^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 2^{1} (i 3)^{3} : 243 i^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} \cdot 2^{0} (i 3)^{4} : 9 i^{4}

= 16 + 323 i + 72 i^{2} + 243 i^{3} + 9 i^{4}

Simplify

16 + 323 i + 72 i^{2} + 243 i^{3} + 9 i^{4} : 83 i - 47

= 83 i - 47

Rewrite in standard complex form:

- 47 + 83 i

= - 47 + 83 i

(x^{3} + 3) (3 - x^{3})

- \frac{s e ^{- \frac{a}{5}}}{10 ( s ^{2} + 1 ) ( s + 4 ) ( s + 5 )}

\frac{( 1 - i ) ^{13}}{64}

\frac{1}{s ( a s + 1 ) ( b s + 1 )}

\frac{x ^{2} - 3 x}{( x + 3 ) ^{2}}