Question

Z=(1+2i)/(1-〖(1-i)〗^2,then arg(z)=?

Answer 1

Expanding out the denominator,\[\Large\rm z=\frac{1+2\mathcal i}{1-(1-2\mathcal i -1)}\]Which simplifies to,\[\Large\rm z=\frac{1+2\mathcal i}{1+2\mathcal i}=1\]z=1 is along the real axis, yes?
So our angle is zero, but if it's not the principle argument (usually denoted with a lowercase a) then we include the 2kpi.\[\Large\rm \arg(z)=0+2k \pi\]Something like that I believe..