Rozwiązanie:
Obliczamy najpierw wartość liczby pod modułem:
[tex]$\frac{(1-2i)(2+3i)}{2-i} =\frac{2+3i-4i+6}{2-i} =\frac{8-i}{2-i} =\frac{(8-i)(2+i)}{(2-i)(2+i)} =[/tex]
[tex]$=\frac{16+8i-2i+1}{4+1} =\frac{17+6i}{5} =\frac{17}{5} +\frac{6}{5}i [/tex]
Zatem:
[tex]$\Big|\frac{17}{5} +\frac{6}{5}i \Big|=\sqrt{\Big(\frac{17}{5}\Big)^{2}+\Big(\frac{6}{5}\Big)^{2}} =\sqrt{\frac{289}{25}+\frac{36}{25} } =\sqrt{\frac{325}{25} } =\sqrt{13} [/tex]
