# Partial Derivatives for Functions f(z) of a Complex Variable.

1. Jul 23, 2010

### Bacle

Hi, Everyone:

I was never clear n this point: given that z is a single complex variable,
how/why does it make sense to talk about z having partial derivatives.?

I mean, if we are given, say, f(x,y); R<sup>2</sup> -->R<sup>n</sup>
then it makes sense to talk about f<sub>x</sub> and f<sub>y</sub>, since
x and y are different variables. But , in f(z), z is a single variable, so there are no

Is it the case that a function of a complex variable z is also a function of two complex
variables.?. If not, is there a formal/theoretical argument to support this use.?
Thanks.

2. Jul 23, 2010

### Anthony

If we were to use the natural identification $$\mathbf{R}^2 \simeq \mathbf{C}$$, then you can write a complex valued function on $$\mathbf{R}^2$$ via:

$$f(x,y) = f\left( \frac{z+\bar z}{2}, \frac{z-\bar z}{2\mathrm{i}}\right) \equiv F(z,\bar z)$$

The notion of complex differentiability is determined by the Cauchy-Riemann equations: $$\partial F/\partial \bar z =0$$.

3. Jul 23, 2010

### snipez90

Anthony's argument supports the idea that analytic functions are true functions of a complex variable as opposed to simply complex functions of two real variables.

As to your very first question, I'm not sure what the problem is. We can write a complex function of a complex variable as f(x + iy) = u(x,y) + iv(x,y), where u and v are real functions. Partial derivatives then work exactly as they do in real analysis...

4. Jul 23, 2010

### Bacle

But you do say that f(z) is a function of two real variables, through:

f(z)=u(x,y)+iv(x,y) .?

Only then does it make sense to take partials f_x and f_y.

But if f is a function of z as well as a function of x,y, is there
some functional dependence between z and x and z and y.?

My point is that if an analytic function depends only on a single
complex variable, then the argument z has only 1 "part" , so that
a partial derivative would not make sense, just like the partial
derivative of a function of a single real variable would not make sense.
(unlike a function f(x,y) , whose argument has two "parts" )

I thought that, re what Anthony said, that we may be using implicitly,
the diffeomorphism between C and R^2 : x+iy -->(x,y) .

But this last (maybe combined with my lack of sleep )seems confusing,
since there are functions f: R^2-->R that are differentiable , while those
same functions f are not differentiable as functions from C-->R (specifically,
all differentiable functions that don't satisfy Cauchy-Riemann ) ; I thought
that if manifolds M,N were diffeomorphic, that meant that every function f:M-->R
is differentiable iff f:N-->R is also differentiable.

Hope I am not too far of.