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4-1.Complex numbers
hard
The maximum value of $|z|$ where z satisfies the condition $\left| {z + \frac{2}{z}} \right| = 2$ is
A
$\sqrt 3 - 1$
B
$\sqrt 3 + 1$
C
$\sqrt 3 $
D
$\sqrt 2 + \sqrt 3 $
Solution
(b)$\left| {z + \frac{2}{z}} \right| = 2 \Rightarrow |z| – \frac{2}{{|z|}} \le 2$$⇒$ $|z{|^2} – 2|z| – 2 \le 0$
$|z| \le \frac{{2 \pm \sqrt {4 + 8} }}{2} \le 1 \pm \sqrt 3 $.
Hence max. value of $|z|$ is $1 + \sqrt 3 $
Standard 11
Mathematics