MCQ
If $\text{z}=\Big(\frac{1+2\text{i}}{1-(1-\text{i})^2}\Big),$ then arg(z) equal:
- A0
- B$\frac{\pi}{2}$
- C$\pi$
- Dnone of these.
Solution:
$\text{z}=\frac{1+2\text{i}}{1-(1-\text{i})^2}$
$\Rightarrow\text{z}=\frac{1+2\text{i}}{1-(1+\text{i}^2-2\text{i})}$
$\Rightarrow\text{z}=\frac{1+2\text{i}}{1-(1-1-2\text{i})}$
$\Rightarrow\text{z}=\frac{1+2\text{i}}{1+2\text{i}}$
$\Rightarrow\text{z}=1$
Since point (1,0) lies on the positive direction of real axis, we have:
arg(z) = 0
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