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If $z$ is any complex number satisfying $|z-3-2 i| \leq 2$, then the maximum value of $|2 z-6+5 i|$ is
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The correct answer is:
5
Given, $|z-3-2 i| \leq 2$
To find minimum of $|2 z-6+5 i|$ or $\quad 2\left|z-3+\frac{5}{2} i\right|$
Using triangle inequality, i.e. ||$z_1|-| z_2|| \leq\left|z_1+z_2\right|$ $\therefore\left|z-3+\frac{5}{2} i\right|$ $=\left|z-3-2 i+2 i+\frac{5}{2} i\right|$ $=\left|(z-3-2 i)+\frac{9}{2} i\right| \geq|| z-3-2 i\left|-\frac{9}{2}\right|$ $\geq\left|2-\frac{9}{2}\right| \geq \frac{5}{2} \Rightarrow\left|z-3+\frac{5}{2} i\right| \geq \frac{5}{2}$ or $\quad|2 z-6+5 i| \geq 5$
To find minimum of $|2 z-6+5 i|$ or $\quad 2\left|z-3+\frac{5}{2} i\right|$
Using triangle inequality, i.e. ||$z_1|-| z_2|| \leq\left|z_1+z_2\right|$ $\therefore\left|z-3+\frac{5}{2} i\right|$ $=\left|z-3-2 i+2 i+\frac{5}{2} i\right|$ $=\left|(z-3-2 i)+\frac{9}{2} i\right| \geq|| z-3-2 i\left|-\frac{9}{2}\right|$ $\geq\left|2-\frac{9}{2}\right| \geq \frac{5}{2} \Rightarrow\left|z-3+\frac{5}{2} i\right| \geq \frac{5}{2}$ or $\quad|2 z-6+5 i| \geq 5$
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