Show that $\displaystyle f(z) = |z|^2$ is differentiable only at z = 0. Hence it is no where analytical.?
$\displaystyle |z|^2=z\cdot \bar{z}=(x+iy)\cdot (x-iy)=x^2+y^2$
Now as $\displaystyle f(z)=u(x,y)+iv(x,y)$
This gives $\displaystyle u(x,y)=x^2+y^2; v(x,y)=0$
Using the cauchy Riemann equaitons
$\displaystyle u_x=2x$ only equals $\displaystyle v_y=0$ at the point x=0
A similar argument shows the same for y so the function is diff at (0,0).