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gerryliyana

  • 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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  1. gerryliyana
    • one year ago
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    @phi @Hero @jhonyy9

  2. Xavier
    • one year ago
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    Use the limit definition of the derivative and you have it approaching different values along different lines

  3. gerryliyana
    • one year ago
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    how to prove it using philosophy of course very deeply??

  4. sirm3d
    • one year ago
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    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 \[f(z)=u(x,y) + v(x,y)i\]

  5. sirm3d
    • one year ago
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    let \[z=x+yi\] \[f(z)= \text{Re } z=x+0i\]

  6. sirm3d
    • one year ago
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    \[u_x=1,v_y=0\] the first condition is not satisfied.

  7. gerryliyana
    • one year ago
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    @sirm3d did you know symmetric tops of molecule??

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