Proving f(z)=zRe(z) is Differentiable at Origin

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SUMMARY

The function f(z) = zRe(z) is differentiable only at the origin, as established through the application of the Cauchy-Riemann equations. By expressing f(z) in terms of its real and imaginary components, we find u(x, y) = x² and v(x, y) = xy. The Cauchy-Riemann equations, u_x = v_y and v_x = -u_y, are satisfied solely at the origin, confirming the differentiability condition. This analysis clarifies the behavior of the function in the complex plane.

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Homework Statement


show that the function f(z)=zRe(z) is only differentiable at the origin.

im completely lost with is, its probably very easy.. but don't know how to start.
f(x+iy)=x+iy*Re(z)=x(x+iy)? idk
 
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I believe you need to use the Cauchy-Riemann equations, u_x = v_y, v_x = -u_y, where
u is the real part of the function and v is the imaginary part, and u_x is the derivative of u with respect to x etc.

As you stated, the function can be written as x(x+iy) = x^2 + ixy, so u = x^2, v = xy. You will find the functions only satisfy the C-R equations at the origin.
 
fredrick08 said:
im completely lost with is, its probably very easy.. but don't know how to start.
f(x+iy)=x+iy*Re(z)=x(x+iy)?

So far so good, now what are the real and imaginary parts, u(x,y) and v(x,y), of this expression? What conditions must the partial derivatives of u(x,y) and v(x,y) satisfy for f(x+iy) to be differentiable?

EDIT: NT123 beat me to it. :smile:
 

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