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