We might suspect the minimum value of is 1
when z=0, as it would be if z were real. But we have to prove it.
Let z = x+iy with x,y real.
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
Consider the function:
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 is the piecewise function:
Thus the minimum value is of is 1.
Edwin