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Popular Álgebra >

simplificar (4+3i)^7

Solução

simplificar

(4 + 3 i)^{7}

Solução

- 16124 - 76443 i

Passos da solução

(4 + 3 i)^{7}

Aplicar o teorema do binômio:

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

a = 4, b = 3 i

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

Expandir a soma

= \frac{7 !}{0 ! ( 7 - 0 ) !} \cdot 4^{7} (3 i)^{0} + \frac{7 !}{1 ! ( 7 - 1 ) !} \cdot 4^{6} (3 i)^{1} + \frac{7 !}{2 ! ( 7 - 2 ) !} \cdot 4^{5} (3 i)^{2} + \frac{7 !}{3 ! ( 7 - 3 ) !} \cdot 4^{4} (3 i)^{3} + \frac{7 !}{4 ! ( 7 - 4 ) !} \cdot 4^{3} (3 i)^{4} + \frac{7 !}{5 ! ( 7 - 5 ) !} \cdot 4^{2} (3 i)^{5} + \frac{7 !}{6 ! ( 7 - 6 ) !} \cdot 4^{1} (3 i)^{6} + \frac{7 !}{7 ! ( 7 - 7 ) !} \cdot 4^{0} (3 i)^{7}

Simplificar

\frac{7 !}{0 ! ( 7 - 0 ) !} \cdot 4^{7} (3 i)^{0} : 16384

Simplificar

\frac{7 !}{1 ! ( 7 - 1 ) !} \cdot 4^{6} (3 i)^{1} : 86016 i

Simplificar

\frac{7 !}{2 ! ( 7 - 2 ) !} \cdot 4^{5} (3 i)^{2} : 193536 i^{2}

Simplificar

\frac{7 !}{3 ! ( 7 - 3 ) !} \cdot 4^{4} (3 i)^{3} : 241920 i^{3}

Simplificar

\frac{7 !}{4 ! ( 7 - 4 ) !} \cdot 4^{3} (3 i)^{4} : 181440 i^{4}

Simplificar

\frac{7 !}{5 ! ( 7 - 5 ) !} \cdot 4^{2} (3 i)^{5} : 81648 i^{5}

Simplificar

\frac{7 !}{6 ! ( 7 - 6 ) !} \cdot 4^{1} (3 i)^{6} : 20412 i^{6}

Simplificar

\frac{7 !}{7 ! ( 7 - 7 ) !} \cdot 4^{0} (3 i)^{7} : 2187 i^{7}

= 16384 + 86016 i + 193536 i^{2} + 241920 i^{3} + 181440 i^{4} + 81648 i^{5} + 20412 i^{6} + 2187 i^{7}

Simplificar

16384 + 86016 i + 193536 i^{2} + 241920 i^{3} + 181440 i^{4} + 81648 i^{5} + 20412 i^{6} + 2187 i^{7} : - 76443 i - 16124

= - 76443 i - 16124

Rewrite in standard complex form:

- 16124 - 76443 i

= - 16124 - 76443 i

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