# Show that f(Z) = Z(hat) is not differentiable.

• Mar 24th 2011, 10:23 AM
davy337
Show that f(Z) = Z(hat) is not differentiable.
Hi ppl, would love the solutuin to "show that f(Z) = Z(hat) is not differentiable"
• Mar 24th 2011, 10:31 AM
FernandoRevilla
Quote:

Originally Posted by davy337
Hi ppl, would love the solutuin to "show that f(Z) = Z(hat) is not differentiable"

I think you mean $\displaystyle f(z)=\bar{z}=x-iy$ . Use the Cauchy Riemann equations for $\displaystyle u=x,v=-y$ .
• Mar 24th 2011, 07:49 PM
mr fantastic
Quote:

Originally Posted by davy337
Hi ppl, would love the solutuin to "show that f(Z) = Z(hat) is not differentiable"

The previous poster has given help. If you need more help, please show all your work and say where exactly you are stuck.
• Mar 25th 2011, 04:08 AM
HallsofIvy
Or you could use the reasoning that leads to the Cauchy-Riemann equations. Let z= x+ iy where x and y are real numbers. Then $\displaystyle \overline{z}= x- iy$. To find the derivate at, say $\displaystyle z_0= x_0+ iy_0$ we form the "difference quotient" $\displaystyle \frac{f(z)- f(z_0)}{z- z_0}=\frac{x- iy- (x_0- iy_0)}{x+ iy- (x_0+ iy_0)}= \frac{x- x_0+ i(-y- y_0)}{x-x)+ i(y- y_0)}$ and take the limit as z goes $\displaystyle z_0= x_0+ iy_0$. Since that is a two dimensional limit, in order that the derivative exist, the limit must be the same as we approach from any direction.

In particular, what do we get if we approach $\displaystyle (x_0, y_0)$ along the line $\displaystyle (x, y_0)$ ?

What do we get if we approach $\displaystyle (x_0, y_0)$ along the line $\displaystyle (x_0, y)$? Are those the same?
• Mar 25th 2011, 01:28 PM
mr fantastic
Quote:

Originally Posted by FernandoRevilla
I think you mean $\displaystyle f(z)=\bar{z}=x-iy$ . Use the Cauchy Riemann equations for $\displaystyle u=x,v=-y$ .

There is also a useful (but perhaps not commonly known) Cauchy-Riemann relation for when you have a function of $\displaystyle z$ and $\displaystyle \overline{z}$:

$\displaystyle \displaystyle \frac{\partial f}{\partial \overline{x}} = 0$.