(B) We have, $|z| = \left|z-\frac{3}{z}+\frac{3}{z}\right|$Using the triangle inequality, $|z| \leq \left|z-\frac{3}{z}\right| + \left|\frac{3}{z}\rig

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(B) We have, $|z| = \left|z-\frac{3}{z}+\frac{3}{z}\right|$Using the triangle inequality, $|z| \leq \left|z-\frac{3}{z}\right| + \left|\frac{3}{z}\right|$Given $\left|z-\frac{3}{z}\right| = 2$, we get $|z| \leq 2 + \frac{3}{|z|}$Multiplying by $|z|$ (since $|z| > 0$), we have $|z|^2 \leq 2|z| + 3$$|z|^2 - 2|z| - 3 \leq 0$$(|z|-3)(|z|+1) \leq 0$Since $|z| \geq 0$, we must have $|z| \leq 3$Thus, the maximum value of $|z|$ is $3$.

Class 11 Mathematics 1 Complex Numbers and Quadratic Equations: Complex Numbers Mix Examples-Complex numbers

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Find the maximum value of $|z|$ when $\left|z-\frac{3}{z}\right|=2,$ where $z$ is a complex number.

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A
$1+\sqrt{3}$
B
$3$
C
$1+\sqrt{2}$
D
$1$

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