A) \[\sqrt{3}-1\]
B) \[\sqrt{3}+1\]
C) \[\sqrt{3}\]
D) \[\sqrt{2}+\sqrt{3}\]
Correct Answer: B
Solution :
We have \[\left| z \right|=\left| z+\frac{2}{z}-\frac{2}{z} \right|\le \left| z+\frac{2}{z} \right|+\frac{2}{\left| \,z\, \right|}\] \[[\because \,\,\left| {{z}_{1}}\pm {{z}_{2}} \right|\le \left| {{z}_{1}} \right|+\left| {{z}_{2}} \right|]\] which implies \[\left| z \right|\le 2+\frac{2}{\left| \,z\, \right|}\,\,\,\,\,\,\Rightarrow \,\,{{\left| \,z\, \right|}^{2}}\le 2\left| \,z\, \right|+2\] \[\Rightarrow \,\,\,\,|z{{|}^{2}}-\,2\,\,\left| z \right|\,\,+\,1\le 1+2\,\,\,\Rightarrow \,\,{{(\left| \,z\, \right|-1)}^{2}}\,\le 3\] \[\Rightarrow \,\,\,\sqrt{3}\le \,|z|-1\le \,\,\sqrt{3}\,\,\Rightarrow \,\,1-\sqrt{3}\le \,\,\left| \,z\, \right|\,\le \,\,1+\sqrt{3}\] Thus, the maximum value of \[\left| z \right|\,\,is\,\,1+\sqrt{3}\]
You need to login to perform this action.
You will be redirected in
3 sec
