# Technical question in multi-variable differentiation

1. Feb 2, 2008

### jjou

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

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...)

2. Feb 2, 2008

### nuclearrape66

no youre not allowed.

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

3. Feb 2, 2008

### quasar987

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

4. Feb 3, 2008

### HallsofIvy

Staff Emeritus
Perhaps you could but that has nothing to do with the derivative. My question is why on earth would you even consider that $u^2_x+ v^2_y= 0$?

5. Feb 3, 2008

### jjou

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

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

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

Will have to rethink the proof...