Showing $f(z)=|z|^{1/2}z$ is differentiable at $z=0$ but not holomorphic. Given  $f(z)=|z|^{1/2}z$, it is obviously complex differentiable at $z=0$ because $$f'(0)=\lim_{z\rightarrow 0}\frac{|z|^{1/2}z}{z}=0$$ I have done similar examples with $|z|^2$, but the $|z|^{1/2}$ is throwing me off because you cannot simply expand it like you would with $|z|^2$. How can I show that it is not holomorphic at $z=0$?
Also, what would be the set of all $z$ such that $f$ is complex differentiable? I assume I have to use Cauchy-Riemann equation, but again, I'm having trouble working with the $|z|^{1/2}$
 A: Holomorphicity at $0$ means that it's complex-differentiable in a neighbourhood
of $0$. But it's not complex-differentiable other than at $0$.
For the C-R equations, one has $f(x+iy)=u+iv$
where
$$u=x(x^2+y^2)^{1/4}$$
and
$$v=y(x^2+y^2)^{1/4}.$$
Then
$$u_x=(x^2+y^2)^{1/4}+\frac{x^2}{2(x^2+y^2)^{3/4}}
=\frac{3x^2+2y^2}{2(x^2+y^2)^{3/4}}$$
and similarly,
$$v_y=\frac{2x^2+3y^2}{2(x^2+y^2)^{3/4}}.$$
These are different at nonzero points on the $x$-axis, so $f$ isn't
complex-differentiable there, and so not in any neighbourhood of $0$.
A: Substistute $z = x+iy$
Separate your function into real parts and imaginary parts.
$|z|^\frac12$ is strictly real
$|z| = (x^2 + y^2)^\frac 12\\
|z|^{\frac 12} = (x^2 + y^2)^\frac 14\\
|z|^{\frac 12}z = (x^2 + y^2)^\frac 14(x+iy)$
$f(x+iy) = u+iv\\
u = (x^2+y^2)^\frac 14x\\
v = (x^2+y^2)^\frac 14y$
Do these satisfy they Cauchy-Reimann equations?
$\frac {\partial u}{\partial x} = \frac {\partial v}{\partial y}\\
\frac {\partial u}{\partial y} = -\frac {\partial v}{\partial x}$
