Nombres complexes

Sachant que \(j^{30} = \left(j^{2}\right)^{15} = \left(-1\right)^{15} = -1\)

et \(j^{19} = j^{18}j = \left(j^{2}\right)^{9}j = -j\)

d'où

\(\begin{array}{llll}\underline{z}_{2} &= \frac{-3+j}{2j-1} \\\\ &=\frac{(-3+j)(-2j-1)}{(2j-1)(-2j-1)} \\\\&=\frac{5+5j}{5} \end{array}\)

donc

\(\underline{z}_{2} = 1+j\) ( 3 points )