A) \[\sqrt{3}-1\]
B) \[\sqrt{3}\]
C) \[\sqrt{3}+1\]
D) \[\sqrt{2}+\sqrt{3}\]
Correct Answer: C
Solution :
We have, \[|z|=\left| z+\frac{2}{z}-\frac{2}{z} \right|\] \[\le \left| 2+\frac{2}{z} \right|+\left| \frac{-2}{z} \right|\] [using triangle inequality] \[\therefore \] \[|z|\le 2+\frac{2}{|z|}\] \[\left[ \because \left| z+\frac{2}{z} \right|=2(given) \right]\] \[\Rightarrow \] \[|z{{|}^{2}}-2|z|-2\le 0\] \[\Rightarrow \] \[(|z|-1+\sqrt{3})(|z|-1-\sqrt{3})\le 0\] \[\Rightarrow \] \[1-\sqrt{3}\le |z|\le 1+\sqrt{3}\] Thus, the maximum value of\[|z|\]is\[1+\sqrt{3}\].
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