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

[thesaruman]

Homework Statement

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.

Homework Equations

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

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?

 

[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.

 

[thesaruman]

Thanks, very much.