JEE Advance - Mathematics (2020 - Paper 1 Offline - No. 9)
Let S be the set of all complex numbers z
satisfying |z2 + z + 1| = 1. Then which of the following statements is/are TRUE?
satisfying |z2 + z + 1| = 1. Then which of the following statements is/are TRUE?
$$\left| {z + {1 \over 2}} \right|$$ $$ \le $$ $${{1 \over 2}}$$ for all z$$ \in $$S
|z| $$ \le $$ 2 for all z$$ \in $$S
$$\left| {z + {1 \over 2}} \right|\, \ge {1 \over 2}$$ for all z$$ \in $$S
The set S has exactly four elements
Explanation
It is given that the complex number satisfying
$$|{z^2} + z + 1| = 1 \Rightarrow \left| {{{\left( {z + {1 \over 2}} \right)}^2} + {3 \over 4}} \right| = 1$$
$$ \because $$ $$|{z_1} - {z_2}|\, \ge \,||{z_1}| - |{z_2}||$$
$$ \therefore $$ $$\left| {{{\left( {z + {1 \over 2}} \right)}^2} - \left( { - {3 \over 4}} \right)} \right|\, \ge \,\left| {{{\left| {z + {1 \over 2}} \right|}^2} - \left| { - {3 \over 4}} \right|} \right|$$
$$ \Rightarrow \left| {{{\left| {z + {1 \over 2}} \right|}^2} - {3 \over 4}} \right|\, \le \,1$$
$$ \Rightarrow - 1 \le \,{\left| {z + {1 \over 2}} \right|^2} - {3 \over 4}\, \le \,1$$
$$ \Rightarrow - {1 \over 4}\, \le \,{\left| {z + {1 \over 2}} \right|^2}\, \le \,{7 \over 4}$$
$$ \Rightarrow 0 \le \,{\left| {z + {1 \over 2}} \right|^2}\, \le \,{7 \over 4}$$ {$$ \because $$ |z| $$ \ge $$ 0}
$$ \Rightarrow 0 \le \,{\left| {z + {1 \over 2}} \right|^2}\, \le \,{{\sqrt 7 } \over 2}$$ ....(i)
$$ \because $$ $$ \Rightarrow \,\left| {z + {1 \over 2}} \right|\, \le \,{{\sqrt 7 } \over 2}$$
$$ \because $$ $$|{z_1} + {z_2}|\, \ge \,||{z_1}| - |{z_2}||$$
$$ \therefore $$ $$\left| {z + {1 \over 2}} \right|\, \ge \,\left| {|z| - {1 \over 2}} \right|$$
$$ \Rightarrow \left| {|z| - {1 \over 2}} \right|\, \le \,\left| {z + {1 \over 2}} \right| \le \,{{\sqrt 7 } \over 2}$$
$$ \Rightarrow - {{\sqrt 7 } \over 2}\, \le \,|z| - {1 \over 2} \le {{\sqrt 7 } \over 2}$$
$$ \Rightarrow {{1 - \sqrt 7 } \over 2} \le \,|z|\, \le {{\sqrt 7 + 1} \over 2}$$
$$ \Rightarrow |z|\, \le \,{{1 + \sqrt 7 } \over 2}$$, $$ \therefore $$ |z| $$ \le $$ 2
$$ \because $$ $$\left| {{{\left( {z + {1 \over 2}} \right)}^2} + {3 \over 4}} \right|\, \le \,\left| {{{\left| {z + {1 \over 2}} \right|}^2} + {3 \over 4}} \right|$$
$$ \Rightarrow 1 \le \left| {{{\left| {z + {1 \over 2}} \right|}^2} + {3 \over 4}} \right| \Rightarrow {\left| {z + {1 \over 2}} \right|^2} + {3 \over 4} \ge 1$$
$$ \Rightarrow {\left| {z + {1 \over 2}} \right|^2} \ge {1 \over 4} \Rightarrow \left| {z + {1 \over 2}} \right|\, \ge {1 \over 2}$$ ....(ii)
from Eqs. (i) and (ii), we get
$${1 \over 2} \le \left| {z + {1 \over 2}} \right|\, \le {{\sqrt 7 } \over 2}$$
$$|{z^2} + z + 1| = 1 \Rightarrow \left| {{{\left( {z + {1 \over 2}} \right)}^2} + {3 \over 4}} \right| = 1$$
$$ \because $$ $$|{z_1} - {z_2}|\, \ge \,||{z_1}| - |{z_2}||$$
$$ \therefore $$ $$\left| {{{\left( {z + {1 \over 2}} \right)}^2} - \left( { - {3 \over 4}} \right)} \right|\, \ge \,\left| {{{\left| {z + {1 \over 2}} \right|}^2} - \left| { - {3 \over 4}} \right|} \right|$$
$$ \Rightarrow \left| {{{\left| {z + {1 \over 2}} \right|}^2} - {3 \over 4}} \right|\, \le \,1$$
$$ \Rightarrow - 1 \le \,{\left| {z + {1 \over 2}} \right|^2} - {3 \over 4}\, \le \,1$$
$$ \Rightarrow - {1 \over 4}\, \le \,{\left| {z + {1 \over 2}} \right|^2}\, \le \,{7 \over 4}$$
$$ \Rightarrow 0 \le \,{\left| {z + {1 \over 2}} \right|^2}\, \le \,{7 \over 4}$$ {$$ \because $$ |z| $$ \ge $$ 0}
$$ \Rightarrow 0 \le \,{\left| {z + {1 \over 2}} \right|^2}\, \le \,{{\sqrt 7 } \over 2}$$ ....(i)
$$ \because $$ $$ \Rightarrow \,\left| {z + {1 \over 2}} \right|\, \le \,{{\sqrt 7 } \over 2}$$
$$ \because $$ $$|{z_1} + {z_2}|\, \ge \,||{z_1}| - |{z_2}||$$
$$ \therefore $$ $$\left| {z + {1 \over 2}} \right|\, \ge \,\left| {|z| - {1 \over 2}} \right|$$
$$ \Rightarrow \left| {|z| - {1 \over 2}} \right|\, \le \,\left| {z + {1 \over 2}} \right| \le \,{{\sqrt 7 } \over 2}$$
$$ \Rightarrow - {{\sqrt 7 } \over 2}\, \le \,|z| - {1 \over 2} \le {{\sqrt 7 } \over 2}$$
$$ \Rightarrow {{1 - \sqrt 7 } \over 2} \le \,|z|\, \le {{\sqrt 7 + 1} \over 2}$$
$$ \Rightarrow |z|\, \le \,{{1 + \sqrt 7 } \over 2}$$, $$ \therefore $$ |z| $$ \le $$ 2
$$ \because $$ $$\left| {{{\left( {z + {1 \over 2}} \right)}^2} + {3 \over 4}} \right|\, \le \,\left| {{{\left| {z + {1 \over 2}} \right|}^2} + {3 \over 4}} \right|$$
$$ \Rightarrow 1 \le \left| {{{\left| {z + {1 \over 2}} \right|}^2} + {3 \over 4}} \right| \Rightarrow {\left| {z + {1 \over 2}} \right|^2} + {3 \over 4} \ge 1$$
$$ \Rightarrow {\left| {z + {1 \over 2}} \right|^2} \ge {1 \over 4} \Rightarrow \left| {z + {1 \over 2}} \right|\, \ge {1 \over 2}$$ ....(ii)
from Eqs. (i) and (ii), we get
$${1 \over 2} \le \left| {z + {1 \over 2}} \right|\, \le {{\sqrt 7 } \over 2}$$
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