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Ed Quanta
Oct16-04, 01:04 PM
When I use d, I am referring to a partial derivative here.

So where w(z)=u(x,y) + iv(x,y), and the derivative of w(z) exists, I have shown that

(du/dx)(du/dy) + (dv/dx)(dv/dy) = 0

But I have to give a geometric interpretation of this which is somewhat confusing to me. I am not sure what do here. Should I start by constructing vectors normal to the curve u(x,y)=c1 and v(x,y)=c2? And if so, how do I do this? Thanks for reading and wasting your time on me.

Inquisitive_Mind
Oct16-04, 10:20 PM
Yes, that is correct. You may consider the vectors normal to the curve u(x,y)=c1 and v(x,y)=c2. To construct normal vectors you just take the gradients of the functions u(x,y) and v(x,y). The condition (du/dx)(du/dy) + (dv/dx)(dv/dy) = 0 means that the two sets of normal vectors are orthogonal. What does this say about the curves u(x,y)=c1 and v(x,y)=c2 themselves?