CR equations and differentiability

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Homework Statement


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

Homework Equations



Cauchy-Riemann Equations

The Attempt at a Solution



I calculated the partial derivatives,

u_{x} = 2x
v_{y} = 2y
u_{y} = 0
v_{x} = 0

Then said that for the CR equations to hold,

u_{x}=v_{y} therefore y=x
and
u_{y}=-v_{x} therefore 0=0

Then becuase the partial derivatives are continuous for all x,y, f(z) is differentiable along y=x

f(z) is nowhere analytic because an arbitrarily small open disk centered at any point on the line y=x 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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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