Question 888090
<pre>
We might suspect the minimum value of {{{abs(z)+abs(z-1)}}} is 1
when z=0, as it would be if z were real.  But we have to prove it.

{{{abs(z)+abs(z-1)}}}

Let z = x+iy with x,y real.

{{{sqrt(x^2+y^2)+abs((x+iy)-1)}}}

{{{sqrt(x^2+y^2)+abs(x+iy-1)}}}

{{{sqrt(x^2+y^2)+abs(x-1+iy)}}}

{{{sqrt(x^2+y^2)+sqrt((x-1)^2+y^2)}}}

Clearly if this has minimum value, then y=0,
since any other value of y would make it larger.

So we must find the minimum value of

{{{sqrt(x^2)+sqrt((x-1)^2)}}}

{{{abs(x)+abs(x-1)}}}

Consider the function:

{{{"f(x)"}}}{{{""=""}}}{{{abs(x)+abs(x-1)}}}

We consider six cases:

Case 1:  x=0    Then f(x) = f(0) = 0+1 = 1

Case 2:  x=1    Then f(x) = f(1) = 1+0 = 1

Case 3:  x>0 and x-1>0  which implies x>1

Then f(x) = x+x=1 = 2x-1

Case 4:  x>0 and x-1<0  which implies 0<x<1

Then f(x) = x-(x-1) = 1

Case 5:  x<0 and x-1>0  which is impossible

Then f(x) = x+x=1 = 2x-1

Case 5:  x<0 and x-1>0  which is impossible

Case 6:  x<0 and x-1<0  which implies x<0

Then f(x) = -x-(x-1) = x-x+1 = -2x+1

So {{{"f(x)"}}}{{{""=""}}}{{{abs(x)+abs(x-1)}}} is the piecewise function:

{{{matrix(1,3,"f(x)",""="",system(matrix(3,3,-2x+1,if,x<0,
                                        1,  if,0<=x<1,
                                       2x-1,if,x>=1)))}}}   {{{graph(100,400/3,-1,2,-1,3,abs(x)+abs(x-1))}}}

Thus the minimum value is of {{{abs(z)+abs(z-1)}}} is 1.

Edwin</pre>