Question

# If $z$ is any complex number satisfying $\left|z-3-2i\right|\le 2$, then the minimum value of $\left|2z-6+5i\right|$ is

A

$2$

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B

$1$

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C

$3$

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D

$5$

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Solution

## The correct option is D $5$Explanation for the correct option.Find the minimum value of $\left|2z-6+5i\right|$.The expression $\left|2z-6+5i\right|$ can be written as$\begin{array}{rcl}\left|2z-6+5i\right|& =& 2\left|z-3+\frac{5}{2}i\right|\\ & =& 2\left|z-3-2i+2i+\frac{5}{2}i\right|\\ & =& 2\left|\left(z-3-2i\right)+\frac{9}{2}i\right|\end{array}$Now, for two complex number ${z}_{1}$ and ${z}_{2}$ the triangle inequality can be given as: $\left|\left|{z}_{1}\right|-\left|{z}_{2}\right|\right|\le \left|{z}_{1}+{z}_{2}\right|$.So the minimum value of $\begin{array}{rcl}& & 2\left|\left(z-3-2i\right)+\frac{9}{2}i\right|\end{array}$ is given as:$\begin{array}{rcl}2\left|\left(z-3-2i\right)+\frac{9}{2}i\right|& =& 2\left|\left|\left(z-3-2i\right)\right|-\left|\frac{9}{2}i\right|\right|\\ & =& 2\left|2-\frac{9}{2}\right|\left[|z-3-2i|\mathbf{\le }\mathbf{2}\right]\\ & =& 2\left|-\frac{5}{2}\right|\\ & =& 2×\frac{5}{2}\\ & =& 5\end{array}$So, the minimum value of $\left|2z-6+5i\right|$ is $5$.Hence, the correct option is D.

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