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Popular Álgebra >

simplificar (4+3i)^5

Solución

simplificar

(4 + 3 i)^{5}

Solución

- 3116 - 237 i

Pasos de solución

(4 + 3 i)^{5}

Aplicar el teorema del binomio:

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

a = 4, b = 3 i

= i = 0 \sum 5 (i 5) \cdot 4^{(5 - i)} (3 i)^{i}

Expandir sumatorio

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

Simplificar

\frac{5 !}{0 ! ( 5 - 0 ) !} \cdot 4^{5} (3 i)^{0} : 1024

Simplificar

\frac{5 !}{1 ! ( 5 - 1 ) !} \cdot 4^{4} (3 i)^{1} : 3840 i

Simplificar

\frac{5 !}{2 ! ( 5 - 2 ) !} \cdot 4^{3} (3 i)^{2} : 5760 i^{2}

Simplificar

\frac{5 !}{3 ! ( 5 - 3 ) !} \cdot 4^{2} (3 i)^{3} : 4320 i^{3}

Simplificar

\frac{5 !}{4 ! ( 5 - 4 ) !} \cdot 4^{1} (3 i)^{4} : 1620 i^{4}

Simplificar

\frac{5 !}{5 ! ( 5 - 5 ) !} \cdot 4^{0} (3 i)^{5} : 243 i^{5}

= 1024 + 3840 i + 5760 i^{2} + 4320 i^{3} + 1620 i^{4} + 243 i^{5}

Simplificar

1024 + 3840 i + 5760 i^{2} + 4320 i^{3} + 1620 i^{4} + 243 i^{5} : - 237 i - 3116

= - 237 i - 3116

Escribe en forma compleja estándar:

- 3116 - 237 i

= - 3116 - 237 i

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