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Popular Álgebra >

simplificar (x+iy)^5

Solución

simplificar

(x + i y)^{5}

Solución

(x^{5} - 10 x^{3} y^{2} + 5 x y^{4}) + (5 x^{4} y - 10 x^{2} y^{3} + y^{5}) i

Pasos de solución

(x + i y)^{5}

Aplicar el teorema del binomio:

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

a = x, b = i y

= i = 0 \sum 5 (i 5) x^{(5 - i)} (i y)^{i}

Expandir sumatorio

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

Simplificar

\frac{5 !}{0 ! ( 5 - 0 ) !} x^{5} (i y)^{0} : x^{5}

Simplificar

\frac{5 !}{1 ! ( 5 - 1 ) !} x^{4} (i y)^{1} : 5 i x^{4} y

Simplificar

\frac{5 !}{2 ! ( 5 - 2 ) !} x^{3} (i y)^{2} : 10 i^{2} x^{3} y^{2}

Simplificar

\frac{5 !}{3 ! ( 5 - 3 ) !} x^{2} (i y)^{3} : 10 i^{3} x^{2} y^{3}

Simplificar

\frac{5 !}{4 ! ( 5 - 4 ) !} x^{1} (i y)^{4} : 5 i^{4} x y^{4}

Simplificar

\frac{5 !}{5 ! ( 5 - 5 ) !} x^{0} (i y)^{5} : i^{5} y^{5}

= x^{5} + 5 i x^{4} y + 10 i^{2} x^{3} y^{2} + 10 i^{3} x^{2} y^{3} + 5 i^{4} x y^{4} + i^{5} y^{5}

10 i^{2} x^{3} y^{2} = - 10 x^{3} y^{2}

10 i^{3} x^{2} y^{3} = - 10 i x^{2} y^{3}

5 i^{4} x y^{4} = 5 x y^{4}

i^{5} = i

= x^{5} + 5 i x^{4} y - 10 x^{3} y^{2} - 10 i x^{2} y^{3} + 5 x y^{4} + i y^{5}

Escribe en forma compleja estándar:

(x^{5} - 10 x^{3} y^{2} + 5 x y^{4}) + (5 x^{4} y - 10 x^{2} y^{3} + y^{5}) i

= (x^{5} - 10 x^{3} y^{2} + 5 x y^{4}) + (5 x^{4} y - 10 x^{2} y^{3} + y^{5}) i

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