# Derivative of f(z) with respect to z* does not exist

1. Dec 10, 2008

### thesaruman

1. The problem statement, all variables and given/known data

The function f(z) is analytic. Show that the derivative of f(z) with respect to z* does not exist unless f(z) is a constant.
Hint: Use chain rule and take x = (z+z*)/2, y = (z-z*)/2.

2. Relevant equations

$$\frac{d f}{d z*} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial z*} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial z*}$$.

3. The attempt at a solution

Well, I used this relation, considering that the analyticity of f guarantees this. I'm not sure of this procedure, but it was the only way i figured out to use the hint of the author. Then, the result was this:

$$\frac{d f}{d z*} = \frac{1}{2} \left( \frac{\partial f}{\partial x} + i \frac{\partial f}{\partial x} \right)$$.

Next, I used another relation which I seriously doubt of:

$$\frac{\partial f}{\partial x} = \frac{d f}{d z} \frac{\partial z}{\partial x} = 1$$.

Analogously, I deduced that

$$\frac{\partial f}{\partial y} = \frac{d f}{d z} \frac{\partial z}{\partial y} = i$$.

With these results, the previous equation becomes:

$$\frac{d f}{d z*} = \frac{1}{2} \left( \frac{d f}{d z} - \frac{d f}{d z} \right) = 0$$.

This result sounds like an absurd to me, and this could be the answer by "reductio ad absurdum" but my hypothesis doesn't seem correct (or rigorous). Someone has any idea?

Last edited: Dec 10, 2008
2. Dec 10, 2008

### Dick

You mean,
$$\frac{d f}{d z*} = \frac{1}{2} \left( \frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y} \right)$$
I'm sure. The rest of your argument is correct. And yes, df/dz*=0. So if df/dz* exists, it must be zero. You can also reach the same conclusion by substituting f=u(x,y)+i*v(x,y) into that relation and using the Cauchy-Riemann equations. It is a little confusing to phrase it this way. I would say f(z*) is analytic only if f is constant.

3. Dec 10, 2008

### thesaruman

Thanks, very much.