What is (sqrt(3)+isqrt(7))^4 ?

The solution to (sqrt(3)+isqrt(7))^4 is -68-16sqrt(21)i

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simplify (sqrt(3)+isqrt(7))^4

simplify

(3 + i 7)^{4}

- 68 - 1621 i

Solution steps

(3 + i 7)^{4}

Apply binomial theorem:

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

a = 3, b = i 7

= i = 0 \sum 4 (i 4) (3)^{(4 - i)} (i 7)^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} (3)^{4} (i 7)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (3)^{3} (i 7)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (3)^{2} (i 7)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (3)^{1} (i 7)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (3)^{0} (i 7)^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (3)^{4} (i 7)^{0} : 9

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (3)^{3} (i 7)^{1} : 1221 i

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (3)^{2} (i 7)^{2} : 126 i^{2}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (3)^{1} (i 7)^{3} : 2821 i^{3}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (3)^{0} (i 7)^{4} : 49 i^{4}

= 9 + 1221 i + 126 i^{2} + 2821 i^{3} + 49 i^{4}

Simplify

9 + 1221 i + 126 i^{2} + 2821 i^{3} + 49 i^{4} : - 1621 i - 68

= - 1621 i - 68

Rewrite in standard complex form:

- 68 - 1621 i

= - 68 - 1621 i

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\frac{\partial}{\partial x} ((b) (3 x + b))

(- 1, i) (6 - 4 i)

0.5 (5 w i + 1) + (5 w i + 1) (39 w i + 1) + 38.38 k

What is (sqrt(3)+isqrt(7))^4 ?
The solution to (sqrt(3)+isqrt(7))^4 is -68-16sqrt(21)i