# Math Help - differentiation

1. ## differentiation

Show that $f(z) = |z|^2$ is differentiable only at z = 0. Hence it is no where analytical.?

2. Originally Posted by SNESAM
Show that $f(z) = |z|^2$ is differentiable only at z = 0. Hence it is no where analytical.?
$|z|^2=z\cdot \bar{z}=(x+iy)\cdot (x-iy)=x^2+y^2$

Now as $f(z)=u(x,y)+iv(x,y)$

This gives $u(x,y)=x^2+y^2; v(x,y)=0$

Using the cauchy Riemann equaitons

$u_x=2x$ only equals $v_y=0$ at the point x=0

A similar argument shows the same for y so the function is diff at (0,0).