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