Technical question in multi-variable differentiation

[jjou]

Let [tex]f(x+iy)=u(x,y)+iv(x,y)[/tex]. Suppose we know [tex]|f|^2=u^2+v^2[/tex] is a constant function. Then we are allowed to say that [tex](u^2+v^2)_x=(u^2+v^2)_y=0[/tex]. But are we allowed to differentiate u by x and v by y? IE, are we allowed to make the following statement:
[tex](u^2)_x+(v^2)_y=0[/tex]

I'm guessing 'no', but I'm not too sure why. Intuitively, I would guess that you could change u and v in such a way that those changes balance each other out? (Very unclear way to say it...)

 

[nuclearrape66]

no youre not allowed.

add the functions together and differentiate it with respect to whatever subscipt it is

 

[quasar987]

Not you're not allowed. Consider for instance the functions u²=2x+3y and v²=-2x-3y. Then [tex](u^2+v^2)_x=(u^2+v^2)_y=0[/tex], but [tex](u^2)_x+(v^2)_y=-1[/tex]

 

[HallsofIvy]

Let [tex]f(x+iy)=u(x,y)+iv(x,y)[/tex]. Suppose we know [tex]|f|^2=u^2+v^2[/tex] is a constant function. Then we are allowed to say that [tex](u^2+v^2)_x=(u^2+v^2)_y=0[/tex]. But are we allowed to differentiate u by x and v by y? IE, are we allowed to make the following statement:
[tex](u^2)_x+(v^2)_y=0[/tex]

I'm guessing 'no', but I'm not too sure why. Intuitively, I would guess that you could change u and v in such a way that those changes balance each other out? (Very unclear way to say it...)

Perhaps you could but that has nothing to do with the derivative. My question is why on Earth would you even consider that [itex]u^2_x+ v^2_y= 0[/itex]?

 

[jjou]

My friend and I used that in a complex analysis proof that, for a function [tex]f(x+iy) = u(x,y)+iv(x,y)[/tex] that is holomorphic on an open set, if |f| is constant then f is constant.

If |f| is constant, then [tex]|f|^2 = u^2+v^2[/tex] is constant. Then the derivatives [tex](u^2+v^2)_x[/tex] and [tex](u^2+v^2)_y = 0[/tex].

We then also used the fact that [tex](u^2)_x+(v^2)_y = 0[/tex] and manipulated the three equations using the Cauchy-Riemann equations to show that all partials were equal to zero ([tex]u_x = u_y = v_x = v_y = 0[/tex]).

Will have to rethink the proof...