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

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gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0@phi @Hero @jhonyy9

Xavier
 2 years ago
Best ResponseYou've already chosen the best response.0Use the limit definition of the derivative and you have it approaching different values along different lines

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0how to prove it using philosophy of course very deeply??

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.2use 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 \[f(z)=u(x,y) + v(x,y)i\]

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.2let \[z=x+yi\] \[f(z)= \text{Re } z=x+0i\]

sirm3d
 2 years ago
Best ResponseYou've already chosen the best response.2\[u_x=1,v_y=0\] the first condition is not satisfied.

gerryliyana
 2 years ago
Best ResponseYou've already chosen the best response.0@sirm3d did you know symmetric tops of molecule??
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