Question

Find minimum of |z|+|z+2| where z is a complex number​

Answer 1

Answer:

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.