Why is the function f(z) = ¯z not differentiable for any z ∈ C?

[Applejacks]

Homework Statement

Show that f(z) = ¯z is not differentiable for any z ∈ C.

Homework Equations

The Attempt at a Solution

Is it because the Cauchy-Reimann Equations don't hold?

Z (conjugate) = x-iy
u(x,y)=x
v(x,y=-iy

du/dx=1≠dv/dy=-1
du/dy=0≠-dv/dx=0
Edit: Is there another approach? Because the CR Equations is something we learned later on.

 

[Dick]

Homework Statement

Show that f(z) = ¯z is not differentiable for any z ∈ C.

Homework Equations

The Attempt at a Solution

Is it because the Cauchy-Reimann Equations don't hold?

Z (conjugate) = x-iy
u(x,y)=x
v(x,y=-iy

du/dx=1≠dv/dy=-1
du/dy=0≠-dv/dx=0
Edit: Is there another approach? Because the CR Equations is something we learned later on.

Sure. Use the definition of f'(z)=lim h->0 (f(z+h)-f(z))/h. Show the limit is different if you pick h to be real from the limit if you pick h to be imaginary. That's really what the content of the CR equations is.

 

[Applejacks]

I think I get it now.

(f(z+h)-f(z))/h

(conjugate((z+h)-z))/h = h(conjugate)/h

If h=Δx, the ratio equals 1
If h=Δiy, the ratio equals -1.

Since the two approaches don't agree for any z, z(conj) is not analytic anywhere. Correct?

 

[Dick]

I think I get it now.

(f(z+h)-f(z))/h

(conjugate((z+h)-z))/h = h(conjugate)/h

If h=Δx, the ratio equals 1
If h=Δiy, the ratio equals -1.

Since the two approaches don't agree for any z, z(conj) is not analytic anywhere. Correct?

Yep, that's it. That's how you derive CR.