Question:
If $z=\frac{1+2 i}{1-(1-i)^{2}}$, then arg $(z)$ equal
(a) 0
(b) $\frac{\pi}{2}$
(c) π
(d) none of these.
Solution:
(a) 0
Let $z=\frac{1+2 i}{1-(1-i)^{2}}$
$\Rightarrow z=\frac{1+2 i}{1-\left(1+i^{2}-2 i\right)}$
$\Rightarrow z=\frac{1+2 i}{1-(1-1-2 i)}$
$\Rightarrow z=\frac{1+2 i}{1+2 i}$
$\Rightarrow z=1$
Since point $(1,0)$ lies on the positive direction of real axis, we have: $\arg (z)=0$
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