JEE MAIN - Mathematics (2019 - 10th April Morning Slot - No. 5)
If a > 0 and z = $${{{{\left( {1 + i} \right)}^2}} \over {a - i}}$$, has magnitude $$\sqrt {{2 \over 5}} $$, then $$\overline z $$ is equal to :
$$ - {1 \over 5} + {3 \over 5}i$$
$$ - {1 \over 5} - {3 \over 5}i$$
$${1 \over 5} - {3 \over 5}i$$
$$ - {3 \over 5} - {1 \over 5}i$$
Explanation
$$z = {{{{\left( {1 + i} \right)}^2}} \over {a - i}} \times {{a + i} \over {a + i}}$$
$$ \Rightarrow z = {{\left( {1 - 1 + 2i} \right)\left( {a + i} \right)} \over {{a^2} + 1}} = {{2ai - 2} \over {{a^2} + 1}}$$
$$ \Rightarrow \left| z \right| = \sqrt {{{\left( {{{ - 2} \over {{a^2} + 1}}} \right)}^2} + {{\left( {{{2a} \over {{a^2} + 1}}} \right)}^2}} = \sqrt {{{4 + 4{a^2}} \over {{{({a^2} + 1)}^2}}}} $$
$$ \Rightarrow \sqrt {{{4(1 + {a^2})} \over {{{({a^2} + 1)}^2}}}} = {2 \over {\sqrt {{a^2} + 1} }}$$.... (i)
Now given $$\left| z \right| = \sqrt {{2 \over 5}} $$
so $$\sqrt {{2 \over 5}} = {2 \over {\sqrt {1 + {a^2}} }}$$ from equation (i)
By squaring both sides
$$ \Rightarrow {2 \over 5} = {4 \over {1 + {a^2}}}$$
$$ \Rightarrow 1 + {a^2} = 10$$
a2 = 9
$$ \Rightarrow $$ a = $$ \pm $$ 3
$$ \because $$ (a > 0) then a = 3
Now $$\overline z = {{{{\left( {1 - i} \right)}^2}} \over {3 + i}}$$
= $${{\left( {1 + {i^2} - 2i} \right)} \over {3 + i}}$$
= $${{\left( {1 + {i^2} - 2i} \right)\left( {3 - i} \right)} \over {\left( {3 + i} \right)\left( {3 - i} \right)}}$$
= $${{\left( { - 2i} \right)\left( {3 - i} \right)} \over {\left( {9 - {i^2}} \right)}}$$
= $${{\left( { - 2i} \right)\left( {3 - i} \right)} \over {10}}$$
= $${{ - 6i + 2{i^2}} \over {10}}$$
= $${{ - 6i - 2} \over {10}}$$
= $$-{1 \over 5} - {3 \over 5}i$$
$$ \Rightarrow z = {{\left( {1 - 1 + 2i} \right)\left( {a + i} \right)} \over {{a^2} + 1}} = {{2ai - 2} \over {{a^2} + 1}}$$
$$ \Rightarrow \left| z \right| = \sqrt {{{\left( {{{ - 2} \over {{a^2} + 1}}} \right)}^2} + {{\left( {{{2a} \over {{a^2} + 1}}} \right)}^2}} = \sqrt {{{4 + 4{a^2}} \over {{{({a^2} + 1)}^2}}}} $$
$$ \Rightarrow \sqrt {{{4(1 + {a^2})} \over {{{({a^2} + 1)}^2}}}} = {2 \over {\sqrt {{a^2} + 1} }}$$.... (i)
Now given $$\left| z \right| = \sqrt {{2 \over 5}} $$
so $$\sqrt {{2 \over 5}} = {2 \over {\sqrt {1 + {a^2}} }}$$ from equation (i)
By squaring both sides
$$ \Rightarrow {2 \over 5} = {4 \over {1 + {a^2}}}$$
$$ \Rightarrow 1 + {a^2} = 10$$
a2 = 9
$$ \Rightarrow $$ a = $$ \pm $$ 3
$$ \because $$ (a > 0) then a = 3
Now $$\overline z = {{{{\left( {1 - i} \right)}^2}} \over {3 + i}}$$
= $${{\left( {1 + {i^2} - 2i} \right)} \over {3 + i}}$$
= $${{\left( {1 + {i^2} - 2i} \right)\left( {3 - i} \right)} \over {\left( {3 + i} \right)\left( {3 - i} \right)}}$$
= $${{\left( { - 2i} \right)\left( {3 - i} \right)} \over {\left( {9 - {i^2}} \right)}}$$
= $${{\left( { - 2i} \right)\left( {3 - i} \right)} \over {10}}$$
= $${{ - 6i + 2{i^2}} \over {10}}$$
= $${{ - 6i - 2} \over {10}}$$
= $$-{1 \over 5} - {3 \over 5}i$$
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