Question
If $$\left| z \right| = \max \left\{ {\left| {z - 1} \right|,\left| {z + 1} \right|} \right\}$$ then
A.
$$\left| {z + \bar z} \right| = \frac{1}{2}$$
B.
$$z + \bar z = 1$$
C.
$$\left| {z + \bar z} \right| = 1$$
D.
None of these
Answer :
None of these
Solution :
$$\eqalign{
& {\text{If}}\,\,\left| {z - 1} \right| > \left| {z + 1} \right|,{\text{then}}\,{\text{max}}\,\left\{ {\left| {z - 1} \right|,\left| {z + 1} \right|} \right\} = \left| {z - 1} \right| \cr
& \Rightarrow \,{\text{If}}\,\,{\left| z \right|^2} + 1 - z - \bar z > {\left| z \right|^2} + 1 + z + \bar z\,\,{\text{then}}\,\,\left| z \right| = \left| {z - 1} \right| \cr
& \Rightarrow \,{\text{If}}\,\,z + \bar z\,{\text{ < }}\,{\text{0}}\,\,{\text{then}}\,\,{\left| z \right|^2} = {\left| z \right|^2} + 1 - z - \bar z \cr
& \Rightarrow \,{\text{If}}\,\,z + \bar z\,{\text{ < }}\,{\text{0}}\,\,{\text{then}}\,\,z + \bar z = 1, \cr} $$
which is not possible.
$$\eqalign{
& {\text{Again,}}\,\,{\text{If}}\,\,\left| {z + 1} \right| > \left| {z - 1} \right|\,\,{\text{then}}\,\,{\text{max}}\,\left\{ {\left| {z - 1} \right|,\left| {z + 1} \right|} \right\} = \left| {z + 1} \right| \cr
& \Rightarrow \,{\text{If}}\,{\left| z \right|^2} + 1 + z + \bar z > {\left| z \right|^2} + 1 - z - \bar z\,\,{\text{then}}\,\,\left| z \right| = \left| {z + 1} \right| \cr
& \Rightarrow \,{\text{If}}\,\,z + \bar z > 0\,\,{\text{then}}\,\,{\left| z \right|^2} = {\left| z \right|^2} + 1 + z + \bar z \cr
& \Rightarrow \,{\text{If}}\,\,z + \bar z > 0\,\,{\text{then}}\,\,z + \bar z = - 1\,\,{\text{Not}}\,{\text{possible}}\,{\text{again}}. \cr} $$
Therefore the given result cannot hold.