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Question
If \[z = a + ib\] lies in third quadrant, then \[\frac{\bar{z}}{z}\] also lies in third quadrant if
Options
\[a > b > 0\]
\[a < b < 0\]
\[b < a < 0\]
\[b > a > 0\]
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Solution
Since, \[z = a + ib\] lies in third quadrant. \[\Rightarrow a < 0 \text { and } b < 0 . . . . (1)\]
Now,
\[\frac{\bar{z}}{z} = \frac{\bar{{a + ib}}}{a + ib}\]
\[ = \frac{a - ib}{a + ib}\]
\[ = \frac{a - ib}{a + ib} \times \frac{a - ib}{a - ib}\]
\[ = \frac{a^2 + i^2 b^2 - 2abi}{a^2 - i^2 b^2}\]
\[ = \frac{a^2 - b^2 - 2abi}{a^2 + b^2}\]
Since,
\[\frac{\bar{z}}{z}\] also lies in third quadrant.
\[\Rightarrow a^2 - b^2 < 0\]
\[ \Rightarrow (a - b)(a + b) < 0\]
\[ \Rightarrow a - b > 0 \text { and }a + b < 0\]
\[ \Rightarrow a > b . . . . (2)\]
From (1) and (2),
\[b < a < 0\]
Hence, the correct option is (c).
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