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

**davy337** · March 24th 2011, 10:23 AM

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

**FernandoRevilla** · March 24th 2011, 10:31 AM

Originally Posted by davy337

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

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

**mr fantastic** · March 24th 2011, 07:49 PM

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.

**HallsofIvy** · March 25th 2011, 04:08 AM

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 $\overline{z}= x- iy$ . To find the derivate at, say $z_0= x_0+ iy_0$ we form the "difference quotient" $\frac{f(z)- f(z_0)}{z- z_0}=\frac{x- iy- (x_0- iy_0)}{x+ iy- (x_0+ iy_0)}= \frac{x- x_0+ i(-y- y_0)}{x-x)+ i(y- y_0)}$ and take the limit as z goes $z_0= x_0+ iy_0$ . 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 $(x_0, y_0)$ along the line $(x, y_0)$ ?

What do we get if we approach $(x_0, y_0)$ along the line $(x_0, y)$ ? Are those the same?

**mr fantastic** · March 25th 2011, 01:28 PM

Originally Posted by FernandoRevilla

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

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

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

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