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}{\left|z\right|} \left(\because \left|z + \frac{2}{z}\right| = 2\right) $
$\Rightarrow \left|z\right|^{2} - 2\left|z\right| - 2 \le0 $
$ \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}$.