(-1+(2i)/n)^4

 Esercizio

$\left(-1+\frac{2i}{n}\right)^4$

 

Applicare la formula: $\left(a+b\right)^4$$=a^4+4a^3b+6a^2b^2+4ab^3+b^4$, dove $a=-1$, $b=\frac{2i}{n}$ e $a+b=-1+\frac{2i}{n}$

$1+\frac{-8i}{n}+6\left(\frac{2i}{n}\right)^2-4\left(\frac{2i}{n}\right)^3+\left(\frac{2i}{n}\right)^4$

 

Applicare la formula: $\left(\frac{a}{b}\right)^n$$=\frac{a^n}{b^n}$, dove $a=2i$, $b=n$ e $n=2$

$1+\frac{-8i}{n}+6\left(\frac{\left(2i\right)^2}{n^2}\right)-4\left(\frac{2i}{n}\right)^3+\left(\frac{2i}{n}\right)^4$

 

Applicare la formula: $\left(\frac{a}{b}\right)^n$$=\frac{a^n}{b^n}$, dove $a=2i$, $b=n$ e $n=3$

$1+\frac{-8i}{n}+6\left(\frac{\left(2i\right)^2}{n^2}\right)-4\left(\frac{\left(2i\right)^3}{n^3}\right)+\left(\frac{2i}{n}\right)^4$

 

Applicare la formula: $\left(\frac{a}{b}\right)^n$$=\frac{a^n}{b^n}$, dove $a=2i$, $b=n$ e $n=4$

$1+\frac{-8i}{n}+6\left(\frac{\left(2i\right)^2}{n^2}\right)-4\left(\frac{\left(2i\right)^3}{n^3}\right)+\frac{\left(2i\right)^4}{n^4}$

 

Applicare la formula: $\left(ab\right)^n$$=a^nb^n$

$1+\frac{-8i}{n}+6\left(\frac{4i^2}{n^2}\right)-4\left(\frac{8i^3}{n^3}\right)+\frac{16i^4}{n^4}$

$1+\frac{-8i}{n}+6\left(\frac{4i^2}{n^2}\right)-4\left(\frac{8i^3}{n^3}\right)+\frac{16i^4}{n^4}$

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