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

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simplify (x-i)^4

Solution

simplify

(x - i)^{4}

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

Solution steps

(x - i)^{4}

Apply binomial theorem:

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

a = x, b = - i

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

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

= x^{4} - 4 i x^{3} + 6 i^{2} x^{2} - 4 i^{3} x + i^{4}

Simplify

x^{4} - 4 i x^{3} + 6 i^{2} x^{2} - 4 i^{3} x + i^{4} : x^{4} - 4 i x^{3} - 6 x^{2} + 4 i x + 1

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

Rewrite in standard complex form:

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

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

e^{x} (e^{x} + 1)

[(x^{3} - 2 x) \cdot tan (x)]

\frac{21}{x ( x ^{2} + 7 )}

A (- 31) y B (- 2, - 4)

(v - x cos (y))^{2}