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

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

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

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

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

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