anonymous
  • anonymous
prove that the function f(z) = Re(z) does not have a derivative for any value of z
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@phi @Hero @jhonyy9
anonymous
  • anonymous
Use the limit definition of the derivative and you have it approaching different values along different lines
anonymous
  • anonymous
how to prove it using philosophy of course very deeply??

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sirm3d
  • sirm3d
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 \[f(z)=u(x,y) + v(x,y)i\]
sirm3d
  • sirm3d
let \[z=x+yi\] \[f(z)= \text{Re } z=x+0i\]
sirm3d
  • sirm3d
\[u_x=1,v_y=0\] the first condition is not satisfied.
anonymous
  • anonymous
@sirm3d did you know symmetric tops of molecule??

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