Solve the following


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.


(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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