Let complex number z be such that |z - frac{6 | vedclass

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(D) Given $|z - \frac{6}{z}| = 5$. Using the triangle inequality property $||z_1| - |z_2|| \leq |z_1 - z_2|$,we have $|z| - |\frac{6}{z}| \leq |z - \f

Vedclass · Accepted Answer

$6$

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(D) Given $|z - \frac{6}{z}| = 5$. Using the triangle inequality property $||z_1| - |z_2|| \leq |z_1 - z_2|$,we have $|z| - |\frac{6}{z}| \leq |z - \frac{6}{z}|$. Substituting the given value: $|z| - \frac{6}{|z|} \leq 5$. Let $|z| = r$,where $r > 0$. Then $r - \frac{6}{r} \leq 5$. Multiplying by $r$ (since $r > 0$): $r^2 - 6 \leq 5r \Rightarrow r^2 - 5r - 6 \leq 0$. Factoring the quadratic: $(r - 6)(r + 1) \leq 0$. Since $r > 0$,we must have $r \leq 6$. Thus,the maximum value of $|z|$ is $6$.

Let complex number $z$ be such that $|z - \frac{6}{z}| = 5$,then the maximum value of $|z|$ will be -

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