JEE MAIN - Mathematics (2019 - 10th April Evening Slot - No. 17)
If z and w are two complex numbers such that |zw| = 1 and arg(z) – arg(w) = $${\pi \over 2}$$
, then :
$$z\overline w = {{1 - i} \over {\sqrt 2 }}$$
$$\overline z w = i$$
$$z\overline w = {{ - 1 + i} \over {\sqrt 2 }}$$
$$\overline z w = -i$$
Explanation
$$\left| {zw} \right| = 1$$
$$ \Rightarrow $$ $$\left| z \right|\left| w \right| = 1$$
Let $$w = {1 \over r}{e^{i\theta }}$$
then z = $$r{e^{i\left( {\theta + {\pi \over 2}} \right)}}$$
$$\overline z w = {e^{ - i\left( {\theta + {\pi \over 2}} \right)}}.{e^{i\theta }} = {e^{ - i(\pi /2)}} = - i$$
$$z\overline w = {e^{i\left( {\theta + {\pi \over 2}} \right)}}.{e^{ - i\theta }} = {e^{i\pi /2}} = i$$
$$ \Rightarrow $$ $$\left| z \right|\left| w \right| = 1$$
Let $$w = {1 \over r}{e^{i\theta }}$$
then z = $$r{e^{i\left( {\theta + {\pi \over 2}} \right)}}$$
$$\overline z w = {e^{ - i\left( {\theta + {\pi \over 2}} \right)}}.{e^{i\theta }} = {e^{ - i(\pi /2)}} = - i$$
$$z\overline w = {e^{i\left( {\theta + {\pi \over 2}} \right)}}.{e^{ - i\theta }} = {e^{i\pi /2}} = i$$
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