KVPY #16

In the complex plane $|z| = 2$ is represented by a circle (blue) of radius 2 centered at the origin $(0,0)$ , while $|z-1|$ is represented by another circle (red) of radius 2 centered at $(-1,0)$ . $|z-1|$ is maximum when $\theta = \pi$ , when $|z-1|_{max} = |-2-1| = 3$ . Similarly, $|z-2|_{max} = 4$ and $|z-3|_{max} = 5$ . Therefore,

$\implies |z-1|_{max} + |z-2|_{max} + |z-3|_{max} = 3+4+5 = \boxed{12}$

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Algebraic solution

Since $|z|=2$ , $\implies z = 2(\cos \theta + i\sin \theta)$ , where $\theta = \arg(z)$ .

$\begin{aligned} |z-a| & = \sqrt{(2\cos \theta -a)^2 +2^2\sin^2 \theta} & \small \color{#3D99F6} \text{where } a \text{ is a positive constant.} \\ & = \sqrt{4\cos^2 \theta -4a\cos \theta + a^2 +4\sin^2 \theta} \\ & = \sqrt{4+a^2 -4a\cos \theta} \end{aligned}$

This implies that $|z-a|$ is maximum when $\cos \theta = -1$ and:

$\begin{aligned} |z-a|_{max} & = \sqrt{a^2 +4a + 4} = \sqrt{(a+2)^2} = a+2 \end{aligned}$

$\begin{aligned} \implies |z-1|_{max} + |z-2|_{max} + |z-3|_{max} & = 1+2+2+2+3+2 = \boxed{12} \end{aligned}$