Question

If $z=\frac{1+2i}{1-(1-i{)}^2}$, then arg (z) equal
(a) 0
(b) $\frac{\mathrm{\pi }}{2}$
(c) π
(d) none of these.

Answer 1

(a) 0

$\mathrm{Let}z=\frac{1+2i}{1-{\left(1-i\right)}^2}$
$\Rightarrow z=\frac{1+2i}{1-\left(1+i^2-2i\right)}$
$\Rightarrow z=\frac{1+2i}{1-\left(1-1-2i\right)}$
$\Rightarrow z=\frac{1+2i}{1+2i}$
$$\begin{gathered} \Rightarrow z=1 \\ \mathrm{Since}\mathrm{point}(1,0)\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{positive}\mathrm{direction}\mathrm{of}\mathrm{real}\mathrm{axis},\mathrm{we}\mathrm{have}: \\ \arg \left(z\right)=0 \end{gathered}$$