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To find the minimum value of $|z+1|+|z-1|+|z-i|$ where $z\in \Bbb C$. Options:

$(A) \ 2$

$(B) \ 2\sqrt2$

$(C) \ 1+\sqrt3$

$(D) \ \sqrt5$

My logic is the sum will be minimum iff $z\in \Bbb C$ is any one of the three fixed points $1,-1,i$. And by calculation we see that the sum is min when take $z=i$.Is the solution correct?

Know that its not a good solution to the problem....searching for an elegant one...Suggestion reqd..

One can apply Fermat-Torricelli point as given in solution below by Quang Hoang and it is a good solution to the problem geometrically.....but this can be applied only if I know the result...searching a solution of this from known basic results of analysis...

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    $\begingroup$ By symmetry the minimum point has to lie of the imaginary axis. Taking $z=it$ we get the function $2\sqrt{1+t^2} + 1-t$ for $t\in [0,1]$ to minimize. $\endgroup$ – Winther Feb 23 '16 at 23:30
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Hint: Look at the triangle with three vertices: $1$, $-1$, and $i$ on the complex plane. The answer to the question is the Fermat-Torricelli point.

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The minimum is achieved at a point where the angles between the radii to the three points are all equal to 120 degrees, if such a point exists, and one of the points otherwise.

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Hint: A choice among real $z > 0,$ so $$|z+1| + |z-1| + |z-i| = 2z + \sqrt{z^2 + 1}$$ is increazing function.

Answer: (B)

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