The solution to (u+i)^4 is (u^4-6u^2+1)+(4u^3-4u)i

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simplify (u+i)^4

Solution

simplify

(u + i)^{4}

(u^{4} - 6 u^{2} + 1) + (4 u^{3} - 4 u) i

Solution steps

(u + i)^{4}

Apply binomial theorem:

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

a = u, b = i

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

Expand summation

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

Simplify

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

Simplify

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

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} u^{2} i^{2} : 6 i^{2} u^{2}

Simplify

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

Simplify

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

= u^{4} + 4 i u^{3} + 6 i^{2} u^{2} + 4 i^{3} u + i^{4}

6 i^{2} u^{2} = - 6 u^{2}

4 i^{3} u = - 4 i u

i^{4} = 1

= u^{4} + 4 i u^{3} - 6 u^{2} - 4 i u + 1

Rewrite in standard complex form:

(u^{4} - 6 u^{2} + 1) + (4 u^{3} - 4 u) i

= (u^{4} - 6 u^{2} + 1) + (4 u^{3} - 4 u) i

f (- 2) + g (2)

2 \cdot (arcsin (1 - x^{2}) + ∣ arcsin (x) ∣)

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

5 (x + 4 = 25 + 2 x)

A (- 9, - 1, 8) B (- 7, 5, 6)