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

If z is any complex number satisfying |z - 3 - 2i| ≤ 2, then the minimum value of |2z - 6 + 5i| is

$\text{Given: } |z - 3 - 2i| \le 2$

$\text{To find: } \min |2z - 6 + 5i|$

$z = x + iy$

$\text{Centre } 3 + 2i,\ \text{radius } 2$

$\text{Argand (complex) plane}$

$|z - (3 + 2i)| \le 2 \implies A = 3 + 2i,\quad r = 2$

$|2z - 6 + 5i| = |2(z - 3) + 5i|$

$= 2\left|z - 3 + \dfrac{5}{2}i\right|$

$= 2\left|z - \left(3 - \dfrac{5}{2}i\right)\right| = 2\,|z - P|$

$P = 3 - \dfrac{5}{2}i$

$\text{Plan: }1.\ \text{plot the disk and } P$

$2.\ \text{least distance from } P \text{ to the disk}$

$3.\ \text{multiply that distance by } 2$

$CP = |P - A| = \left|\left(3 - \dfrac{5}{2}i\right) - (3 + 2i)\right|$

$= \left| -\dfrac{9}{2}i \right| = \dfrac{9}{2} = 4.5$

$CP > r = 2 \implies P \text{ is outside the disk}$

$\text{Closest point on the disk lies on segment } CP$

$|z - P| \ge CP - |z - A| \ge CP - r$

$\min |z - P| = CP - r = \dfrac{9}{2} - 2 = \dfrac{5}{2}$

$\min |2z - 6 + 5i| = 2 \times \dfrac{5}{2} = 5$

$z_0 = 3 \implies |2(3) - 6 + 5i| = |5i| = 5$

$\boxed{5}$