JEE MAIN - Mathematics (2021 - 20th July Morning Shift - No. 6)
If z and $$\omega$$ are two complex numbers such that $$\left| {z\omega } \right| = 1$$ and $$\arg (z) - \arg (\omega ) = {{3\pi } \over 2}$$, then $$\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)$$ is :
(Here arg(z) denotes the principal argument of complex number z)
(Here arg(z) denotes the principal argument of complex number z)
$${\pi \over 4}$$
$$ - {{3\pi } \over 4}$$
$$ - {\pi \over 4}$$
$${{3\pi } \over 4}$$
Explanation
As $$\left| {z\omega } \right| = 1$$
$$\Rightarrow$$ If $$\left| z \right| = r$$, then $$\left| \omega \right| = {1 \over r}$$
Let $$\arg (z) = \theta $$
$$\therefore$$ $$\arg (\omega ) = \left( {\theta - {{3\pi } \over 2}} \right)$$
So, $$z = r{e^{i\theta }}$$
$$ \Rightarrow \overline z = r{e^{i\theta }}$$
$$\omega = {1 \over r}{e^{i\left( {\theta - {{3\pi } \over 2}} \right)}}$$
Now, consider
$${{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }} = {{1 - 2{e^{i\left( { - {{3\pi } \over 2}} \right)}}} \over {1 + 3{e^{i\left( { - {{3\pi } \over 2}} \right)}}}} = \left( {{{1 - 2i} \over {1 + 3i}}} \right)$$
$$ = {{(1 - 2i)(1 - 3i)} \over {(1 + 3i)(1 - 3i)}} = - {1 \over 2}(1 + i)$$
$$\therefore$$ $$prin\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)$$
$$ = prin\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)$$
$$ = \left( { - {1 \over 2}(1 + i)} \right)$$
$$ = - \left( {\pi - {\pi \over 4}} \right) = {{ - 3\pi } \over 4}$$
So, option (2) is correct.
$$\Rightarrow$$ If $$\left| z \right| = r$$, then $$\left| \omega \right| = {1 \over r}$$
Let $$\arg (z) = \theta $$
$$\therefore$$ $$\arg (\omega ) = \left( {\theta - {{3\pi } \over 2}} \right)$$
So, $$z = r{e^{i\theta }}$$
$$ \Rightarrow \overline z = r{e^{i\theta }}$$
$$\omega = {1 \over r}{e^{i\left( {\theta - {{3\pi } \over 2}} \right)}}$$
Now, consider
$${{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }} = {{1 - 2{e^{i\left( { - {{3\pi } \over 2}} \right)}}} \over {1 + 3{e^{i\left( { - {{3\pi } \over 2}} \right)}}}} = \left( {{{1 - 2i} \over {1 + 3i}}} \right)$$
$$ = {{(1 - 2i)(1 - 3i)} \over {(1 + 3i)(1 - 3i)}} = - {1 \over 2}(1 + i)$$
$$\therefore$$ $$prin\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)$$
$$ = prin\arg \left( {{{1 - 2\overline z \omega } \over {1 + 3\overline z \omega }}} \right)$$
$$ = \left( { - {1 \over 2}(1 + i)} \right)$$
$$ = - \left( {\pi - {\pi \over 4}} \right) = {{ - 3\pi } \over 4}$$
So, option (2) is correct.
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