Question

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

• A. Both Assertion and Reason are individually true and Reason is correct explanation of Assertion.
• B. Both Assertion and Reason are individually true but Reason is not the correct explanation of Assertion.
• C. Assertion is true Reason is false.
• D. Assertion is false Reason is true.

Answer 1

The correct option is B Both Assertion and Reason are individually true but Reason is not the correct explanation of Assertion.

$|z_1|-|z_2|\le |z_1+z_2|\le |z_1|+|z_2|$
Now
$|z|-|\frac{2}{z}|\le |z+\frac{2}{z}|\le |z|+|\frac{2}{z}|$
Or
$|z^2|-2|\le 2|z|\le |z{|}^2+2$
Or
$2|z|\le |z{|}^2+|2|$
$|z{|}^2-2|z|+|2|\ge 0$
Or
$|z{|}^2-2|z|\pm 2\ge 0$
$|z{|}^2-2|z|+2=0$ gives imaginary roots.
Hence
$|z{|}^2-2|z|-2\ge 0$
$|z|=\frac{2\pm \sqrt{12}}{2}$
Or
$|z|=1\pm \sqrt{3}$
Now $1-\sqrt{3}<0$ and $|z|<0$ is not possible.
Hence maximum value is
$|z|=1+\sqrt{3}$