Re: [math-fun] Intermediate value theorem for complex numbers

On 13/02/2013 02:18, rcs@xmission.com wrote:

Suppose F(z) is a continuous function from C->C over the complex numbers, or at least over the unit square. We are silent about the differentiability of F. Suppose that the real part of F, on the vertical edges of the square, has ReF(0+iy)=0 and ReF(1+iy)=0, for 0<=y<=1. On the horizontal edges, the imaginary part of F is similarly constrained: ImF(x+0i)=0 and ImF(x+1i)=0, for 0<=x<=1. (This forces F(z)=0+0i at the four corners of the square.)

I presume that this implies that there is some point z inside the square with F(z) = 0+0i.

No. Consider F(x+iy) = x(1-x) + iy(1-y). -- g