A) \[\max \left| 2{{z}_{1}}+{{z}_{2}} \right|=4\]
B) \[\min \left| {{z}_{1}}-{{z}_{2}} \right|=1\]
C) \[\left| {{z}_{2}}+\frac{1}{{{z}_{1}}} \right|\le 3\]
D) None of these
Correct Answer: D
Solution :
\[\left| 2{{z}_{1}}+{{z}_{2}}\left| \le \right|2{{z}_{1}} \right|+\left| {{z}_{2}} \right|=2\left| {{z}_{1}} \right|+\left| {{z}_{2}} \right|\] |
\[2\times 1+2=4\] |
From the figure, \[\left| {{z}_{1}}-{{z}_{2}} \right|\]is least when \[0,{{z}_{1}},{{z}_{2}}\] are collinear. |
\[\therefore \left| {{z}_{1}}-{{z}_{2}} \right|=1\] |
Again, \[\left| {{z}_{2}}+\frac{1}{{{z}_{1}}} \right|\le \left| {{z}_{2}} \right|+\left| \frac{1}{{{z}_{1}}} \right|=2+\left| \frac{1}{{{z}_{1}}} \right|=2+\frac{1}{1}=3\] |
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