Odpowiedź:
[tex]\frac{13}{5-j} - \frac{13}{5+j} = \frac{13(5+j)}{(5-j)(5+j)} - \frac{13(5-j)}{(5+j)(5-j)} = \frac{45+13j}{25-j^{2} } - \frac{45-13j}{25-j^{2} } = \frac{45-13j-(45-13j)}{25-j^{2} } = \frac{45-13j-45+13j}{25-j^{2} } = \frac{0}{25-j^{2} } = 0[/tex]
[tex]\frac{5-2j}{2+j} - \frac{5+2j}{2-j} = \frac{(5-2j)(2-j)}{(2+j)(2-j)} -\frac{(5+2j)(2+j)}{(2-j)(2+j)} = \frac{10-9j+2j^{2}}{4-j^{2} } - \frac{10+9j+2j^{2} }{4-j^{2} } = \frac{10-9j+2j^{2} -10-9j-2j^{2} }{4-j^{2} } = \frac{-18j}{4-j^{2} }[/tex]
Szczegółowe wyjaśnienie:
