## anonymous 3 years ago prove that the function f(z) = Re(z) does not have a derivative for any value of z

1. anonymous

@phi @Hero @jhonyy9

2. anonymous

Use the limit definition of the derivative and you have it approaching different values along different lines

3. anonymous

how to prove it using philosophy of course very deeply??

4. anonymous

use Cauchy-Riemann conditions for differentiability. $\newcommand {pd}{\partial}\frac{\pd u}{\pd x}=\frac{\pd v}{\pd y},\qquad \frac{\pd u}{\pd y}=-\frac{\pd v}{\pd x}$ given $f(z)=u(x,y) + v(x,y)i$

5. anonymous

let $z=x+yi$ $f(z)= \text{Re } z=x+0i$

6. anonymous

$u_x=1,v_y=0$ the first condition is not satisfied.

7. anonymous

@sirm3d did you know symmetric tops of molecule??