The minimum value of the expression |z|+|z-1| | vedclass

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(B) Let the points in the complex plane be $O(0,0)$,$A(1,0)$,$B(1,1)$,and $C(0,1)$.The expression represents the sum of distances from a point $z$ to

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$2\sqrt{2}$

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(B) Let the points in the complex plane be $O(0,0)$,$A(1,0)$,$B(1,1)$,and $C(0,1)$.The expression represents the sum of distances from a point $z$ to the vertices of a unit square $OABC$.By the triangle inequality,the sum of distances from any point $z$ to the vertices of a convex quadrilateral is minimized at the intersection of its diagonals.The diagonals are $OB$ (connecting $(0,0)$ and $(1,1)$) and $AC$ (connecting $(1,0)$ and $(0,1)$).The intersection point is $z = \frac{1}{2} + \frac{1}{2}i$.The minimum value is the sum of the lengths of the diagonals:$|z-0| + |z-(1+i)| = |\frac{1}{2} + \frac{1}{2}i| + |-\frac{1}{2} - \frac{1}{2}i| = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.$|z-1| + |z-i| = |-\frac{1}{2} + \frac{1}{2}i| + |\frac{1}{2} - \frac{1}{2}i| = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.Total minimum value = $\sqrt{2} + \sqrt{2} = 2\sqrt{2}$.

The minimum value of the expression $|z|+|z-1|+|z-1-i|+|z-i|$,where $z$ is a complex number and $i=\sqrt{-1}$,is

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