Question

Assertion (A) : The maximum value of $|z|$ when $z$ satisfies the condition $\left|z+\frac{2}{z}\right|=2$ is $1+\sqrt{3}$
Reason (R) : $\left|z_1+z_2\right|\le \left|z_1\right|+\left|z_2\right|$.

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A).
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
• C. (A) is true but (R) is false.
• D. (A) is false but (R) is true.

Answer 1

Answer: (B)

We can write
$\left|z\right|=\left|z+\frac{2}{z}-\frac{2}{z}\right|\le \left|z+\frac{2}{z}+\left(-\frac{2}{z}\right)\right|\le \left|z+\frac{2}{z}\right|+\left|\frac{-2}{z}\right|$
$\Rightarrow \left|z\right|<2+\frac{2}{|z|}\left(\because \left|z+\frac{2}{z}\right|=2\right)$
$\Rightarrow {\left|z\right|}^2-2\left|z\right|-2\le 0$
$\Rightarrow \left(\left|z\right|-1+\sqrt{3}\right)\left(\left|z\right|-1\sqrt{3}\right)\le 0$
$\Rightarrow 1-\sqrt{3}\le \left|z\right|\le 1+\sqrt{3}$
$\Rightarrow$ Maximum value of $|z|$ is $1+\sqrt{3}$.