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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Use the limit definition of the derivative and you have it approaching different values along different lines
how to prove it using philosophy of course very deeply??

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

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