Rozwiązanie:
Najpierw obliczmy wartość pod modułem:
[tex]$z=\frac{(1-2i)(2+i)}{2-i} =\frac{2+i-4i+2}{2-i} =\frac{4-3i}{2-i} =\frac{(4-3i)(2+i)}{(2-i)(2+i)} =[/tex]
[tex]$=\frac{8+4i-6i+3}{4+1} =\frac{11-2i}{5} =\frac{11}{5} -\frac{2}{5} i[/tex]
Zatem:
[tex]$\Big|\frac{11}{5}-\frac{2}{5}i \Big|=\sqrt{\Big(\frac{11}{5}\Big)^{2}+\Big(\frac{2}{5}\Big)^{2}} =\sqrt{\frac{121}{25}+\frac{4}{25} } =\sqrt{5} [/tex]
