Rozwiązanie:
[tex]$Im=\frac{(1-i)(2+3i)}{1-2i} =\frac{2+3i-2i+3}{1-2i} =\frac{5+i}{1-2i} =\frac{(5+i)(1+2i)}{(1-2i)(1+2i)} =[/tex]
[tex]$=\frac{5+10i+i-2}{1+4} =\frac{3+11i}{5} =\frac{3}{5} +\frac{11}{5}i [/tex]
Zatem:
[tex]$Im(z)=\frac{11}{5} [/tex]
