Find the minimum value of $|z|+|z-1|+|z-2|; z\in\mathbb C$

Question

Find the minimum value of $|z|+|z-1|+|z-2|; z\in\mathbb C$

I tried this question using many different ways (triangle inequality, geometric interpretation, etc) but I didn't get the correct answer. The correct answer is $2$.

Answer 1

By the triangle inequality, $$\lvert z\rvert+\lvert z-2\rvert=\lvert z\rvert+\lvert 2-z\rvert\geq2.$$ Hence your expression is bounded from below by $2$. Indeed, $2$ is a minimal value because it is attained at $z=1$. If $z\neq 1$ then $\lvert z-1\rvert>0$ and $\lvert z\rvert+\lvert z-2\rvert>2$, so $z=1$ is the unique minimizer.

Answer 2

I will present a different approach.

We let $r\geq 0$ and $z=re^{i\theta}=r(\cos \theta +i\sin \theta) \implies |z|=r$, $|z-1|=\sqrt{(r\cos \theta -1)^2+r^2\sin^2\theta}$, $|z-2|=\sqrt{(r\cos\theta-2)^2+r^2\sin^2\theta}$

Therfore we need to minimize $r+\sqrt{r^2-2r\cos \theta +1}+\sqrt{r^2-4r\cos \theta +4}$

Which clearly occurs at $\theta=0 \implies \cos \theta=1$

Therefore we have to minimize $r+|r-1|+|r-2|$ which is essentially the same expression but here we have $r\geq 0$.

Now assume first that $0 \le r \le 1$, now opening the modulus gives $r+1-r+2-r=3-r$ , we can clearly see that minimum holds for $r=1$.

Now take $1 \leq r \leq 2$, you will again see minimum value comes out to be $2$, at $r=1$.

Finally take the case $r >2$, minimum value comes out to be $3$, but we have to take the absolute minima, thus minima is $2$.

Answer 3

You can use complex geometry in this question.

Consider infinity circles with varying radius centered at $(0,0),(1,0),(2,0)$ in the Argand Plane. Your primary equation says that:
Anywhere be a point z, find the sum of distances from $(0,0),(1,0),(2,0)$ and minimize it.

Now think of this like, if you were given a triangle and you had to find a point such that the sum of distance from each vertex to that point is the least. In our case, the Triangle lies on the $\mathrm{Real-Axis}$, so by analysis, it must be equidistant from $(0,0),(2,0)$ . Hence the answer.