Question

prove that the function f(z) = Re(z) does not have a derivative for any value of z

Answer 1

@phi @Hero @jhonyy9

Answer 2

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

Answer 3

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

Answer 4

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
\[f(z)=u(x,y) + v(x,y)i\]

Answer 5

let \[z=x+yi\]

\[f(z)= \text{Re } z=x+0i\]

Answer 6

\[u_x=1,v_y=0\]
the first condition is not satisfied.

Answer 7

@sirm3d did you know symmetric tops of molecule??