SOLUTION: For any complex number z, the minimum value of |z|+|z-1| is?

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Question 888090: For any complex number z, the minimum value of |z|+|z-1| is?
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
We might suspect the minimum value of abs%28z%29%2Babs%28z-1%29 is 1
when z=0, as it would be if z were real.  But we have to prove it.

abs%28z%29%2Babs%28z-1%29

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

sqrt%28x%5E2%2By%5E2%29%2Babs%28%28x%2Biy%29-1%29

sqrt%28x%5E2%2By%5E2%29%2Babs%28x%2Biy-1%29

sqrt%28x%5E2%2By%5E2%29%2Babs%28x-1%2Biy%29

sqrt%28x%5E2%2By%5E2%29%2Bsqrt%28%28x-1%29%5E2%2By%5E2%29

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%28x%5E2%29%2Bsqrt%28%28x-1%29%5E2%29

abs%28x%29%2Babs%28x-1%29

Consider the function:

%22f%28x%29%22%22%22=%22%22abs%28x%29%2Babs%28x-1%29

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 00  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 %22f%28x%29%22%22%22=%22%22abs%28x%29%2Babs%28x-1%29 is the piecewise function:

   graph%28100%2C400%2F3%2C-1%2C2%2C-1%2C3%2Cabs%28x%29%2Babs%28x-1%29%29

Thus the minimum value is of abs%28z%29%2Babs%28z-1%29 is 1.

Edwin