What is (1+isqrt(3))^5 ?

The solution to (1+isqrt(3))^5 is 16-16sqrt(3)i

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simplify (1+isqrt(3))^5

simplify

(1 + i 3)^{5}

16 - 163 i

Solution steps

(1 + i 3)^{5}

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 5 (i 5) \cdot 1^{(5 - i)} (i 3)^{i}

Expand summation

= \frac{5 !}{0 ! ( 5 - 0 ) !} \cdot 1^{5} (i 3)^{0} + \frac{5 !}{1 ! ( 5 - 1 ) !} \cdot 1^{4} (i 3)^{1} + \frac{5 !}{2 ! ( 5 - 2 ) !} \cdot 1^{3} (i 3)^{2} + \frac{5 !}{3 ! ( 5 - 3 ) !} \cdot 1^{2} (i 3)^{3} + \frac{5 !}{4 ! ( 5 - 4 ) !} \cdot 1^{1} (i 3)^{4} + \frac{5 !}{5 ! ( 5 - 5 ) !} \cdot 1^{0} (i 3)^{5}

Simplify

\frac{5 !}{0 ! ( 5 - 0 ) !} \cdot 1^{5} (i 3)^{0} : 1

Simplify

\frac{5 !}{1 ! ( 5 - 1 ) !} \cdot 1^{4} (i 3)^{1} : 53 i

Simplify

\frac{5 !}{2 ! ( 5 - 2 ) !} \cdot 1^{3} (i 3)^{2} : 30 i^{2}

Simplify

\frac{5 !}{3 ! ( 5 - 3 ) !} \cdot 1^{2} (i 3)^{3} : 303 i^{3}

Simplify

\frac{5 !}{4 ! ( 5 - 4 ) !} \cdot 1^{1} (i 3)^{4} : 45 i^{4}

Simplify

\frac{5 !}{5 ! ( 5 - 5 ) !} \cdot 1^{0} (i 3)^{5} : 93 i^{5}

= 1 + 53 i + 30 i^{2} + 303 i^{3} + 45 i^{4} + 93 i^{5}

Simplify

1 + 53 i + 30 i^{2} + 303 i^{3} + 45 i^{4} + 93 i^{5} : - 163 i + 16

= - 163 i + 16

Rewrite in standard complex form:

16 - 163 i

= 16 - 163 i

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

(x - \frac{1}{x})^{2}

\frac{1}{x ^{2} ( x - 2 )}

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