CR equations and differentiability

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SUMMARY

The function f(z) = x² + i y² is differentiable along the line y = x, as established by the Cauchy-Riemann equations. The calculated partial derivatives are uₓ = 2x, vᵧ = 2y, uᵧ = 0, and vₓ = 0, confirming that uₓ = vᵧ and uᵧ = -vₓ hold true. However, f(z) is not analytic anywhere because any open disk around points on the line y = x contains points where the function is not differentiable. The discussion clarifies that while the Cauchy-Riemann conditions are satisfied, further analysis of directional limits is unnecessary for confirming differentiability along y = x.

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Homework Statement


Where is f(z) differentiable? Analytic?
[tex]f(z) = x^{2} + i y^{2}[/tex]

Homework Equations



Cauchy-Riemann Equations

The Attempt at a Solution



I calculated the partial derivatives,

[tex]u_{x} = 2x[/tex]
[tex]v_{y} = 2y[/tex]
[tex]u_{y} = 0[/tex]
[tex]v_{x} = 0[/tex]

Then said that for the CR equations to hold,

[tex]u_{x}=v_{y}[/tex] therefore [tex]y=x[/tex]
and
[tex]u_{y}=-v_{x}[/tex] therefore [tex]0=0[/tex]

Then because the partial derivatives are continuous for all [tex]x,y[/tex], [tex]f(z)[/tex] is differentiable along [tex]y=x[/tex]

[tex]f(z)[/tex] is nowhere analytic because an arbitrarily small open disk centered at any point on the line [tex]y=x[/tex] will always contain points which are not differentiable.

Is that sufficient to show differentiability? Or am I misapplying the cauchy-riemann conditions?
 
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I suppose what I'm really asking is wether or not I need to look at the limits depending on the direction of approach, or if this is sufficient as is? Help?
 
I think that is good enough. You could explicitly show the derivative limit is independent of direction along x=y, but why? That's what CR are for.
 

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