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

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gerryliyanaBest ResponseYou've already chosen the best response.0
@phi @Hero @jhonyy9
 one year ago

XavierBest ResponseYou've already chosen the best response.0
Use the limit definition of the derivative and you have it approaching different values along different lines
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
how to prove it using philosophy of course very deeply??
 one year ago

sirm3dBest ResponseYou've already chosen the best response.2
use CauchyRiemann 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 \[f(z)=u(x,y) + v(x,y)i\]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.2
let \[z=x+yi\] \[f(z)= \text{Re } z=x+0i\]
 one year ago

sirm3dBest ResponseYou've already chosen the best response.2
\[u_x=1,v_y=0\] the first condition is not satisfied.
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.0
@sirm3d did you know symmetric tops of molecule??
 one year ago
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