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Assertion (A) If $z$ is a complex number such that $|z| \geq 3$, then the least value of $\left|z+\frac{3}{z}\right|$ is 1 .Reason (R) $\left|z_1-z_2\right| \leq\left|z_1\right|+\left|z_2\right|$, for any two complex numbers $z_1, z_2$
The correct option among the following is
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The correct option among the following is
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The correct answer is:
(A) is false but $(R)$ is true.
We have, $|z| \geq 3$
Then, $\left|z+\frac{3}{z}\right| \leq|z|+\left|\frac{3}{z}\right|=\left|z+\frac{3}{z}\right| \leq 3+\left|\frac{3}{z}\right|$
$\therefore$ Least value of $\left|z+\frac{3}{z}\right|$ is not 1
And $\left|z_1-z_2\right| \leq\left|z_1\right|+\left|z_2\right|$ which is true.
Then, $\left|z+\frac{3}{z}\right| \leq|z|+\left|\frac{3}{z}\right|=\left|z+\frac{3}{z}\right| \leq 3+\left|\frac{3}{z}\right|$
$\therefore$ Least value of $\left|z+\frac{3}{z}\right|$ is not 1
And $\left|z_1-z_2\right| \leq\left|z_1\right|+\left|z_2\right|$ which is true.
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