Is the function f(z) = |z|2 differentiable at z0 = a + bi?

[AvgStudent]

f^'(z0)=\stackrel{lim}{x\rightarrow0} \frac{f(z0-z)-f(z0)}{z}

Hi, I'm attempting to use the above equation to show where z0 is not differentiable at some point z0 for the equation

f(z) = |z|²

I was wondering how I could go about doing this?
I tried letting z0 = a + bi, and z = x + yi and got \stackrel{lim}{x\rightarrow0,y\rightarrow0} \frac{2ax+2yb-2ab+x^2+y^2}{x+yi}

I also tired the Cauchy Riemann Equations and let u(x,y) = x^2 + y^2 v(x,y) = 0
f(x,y) = u(x,y) + iv(x,y)
|x+yi| = \sqrt{x^2+y^2}
So 2x = 0, and -2y = 0.
So the equation is differentiable when (x,y) = (0,0)?
Any help is appreciated.

 

[Vargo]

One thing about the Cauchy Riemann equations is that they easily imply that u and v are each harmonic functions...