Question 281 of 455MediumMCQ
If a>0 and Z=a−i(1+i)2 has magnitude 52, then −z is equal to:
Solutions · 1
Official solution
\[z = {{{{\left( {1 + i} \right)}^2}} \over {a - i}} \times {{a + i} \over {a + i}}\]
\[ \Rightarrow z = {{\left( {1 - 1 + 2i} \right)\left( {a + i} \right)} \over {{a^2} + 1}} = {{2ai - 2} \over {{a^2} + 1}}\]
\[ \Rightarrow \left| z \right| = \sqrt {{{\left( {{{ - 2} \over {{a^2} + 1}}} \right)}^2} + {{\left( {{{2a} \over {{a^2} + 1}}} \right)}^2}} = \sqrt {{{4 + 4{a^2}} \over {{{({a^2} + 1)}^2}}}} \]
\[ \Rightarrow \sqrt {{{4(1 + {a^2})} \over {{{({a^2} + 1)}^2}}}} = {2 \over {\sqrt {{a^2} + 1} }}\].... (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 \[\overline z = {{{{\left( {1 - i} \right)}^2}} \over {3 + i}}\]
= \[{{\left( {1 + {i^2} - 2i} \right)} \over {3 + i}}\]
= \[{{\left( {1 + {i^2} - 2i} \right)\left( {3 - i} \right)} \over {\left( {3 + i} \right)\left( {3 - i} \right)}}\]
= \[{{\left( { - 2i} \right)\left( {3 - i} \right)} \over {\left( {9 - {i^2}} \right)}}\]
= \[{{\left( { - 2i} \right)\left( {3 - i} \right)} \over {10}}\]
= \[{{ - 6i + 2{i^2}} \over {10}}\]
= \[{{ - 6i - 2} \over {10}}\]
= \[-{1 \over 5} - {3 \over 5}i\]
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