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\left(2Re(\frac{1}{n+1})Im(n)+2Im(\frac{1}{n+1})Re(n)\right)l=2
La ecuación está en formato estándar.
\frac{\left(2Re(\frac{1}{n+1})Im(n)+2Im(\frac{1}{n+1})Re(n)\right)l}{2Re(\frac{1}{n+1})Im(n)+2Im(\frac{1}{n+1})Re(n)}=\frac{2}{2Re(\frac{1}{n+1})Im(n)+2Im(\frac{1}{n+1})Re(n)}
Divide los dos lados por 2Re(n)Im(\left(n+1\right)^{-1})+2Im(n)Re(\left(n+1\right)^{-1}).
l=\frac{2}{2Re(\frac{1}{n+1})Im(n)+2Im(\frac{1}{n+1})Re(n)}
Al dividir por 2Re(n)Im(\left(n+1\right)^{-1})+2Im(n)Re(\left(n+1\right)^{-1}), se deshace la multiplicación por 2Re(n)Im(\left(n+1\right)^{-1})+2Im(n)Re(\left(n+1\right)^{-1}).
l=\frac{1}{Re(\frac{1}{n+1})Im(n)+Im(\frac{1}{n+1})Re(n)}
Divide 2 por 2Re(n)Im(\left(n+1\right)^{-1})+2Im(n)Re(\left(n+1\right)^{-1}).