Differentiability of the function $f(z)=|z|^p z$($p>0$). Suppose $p>0$ in a real number, is the function $f (z) =|z|^p z$ a differentiable function? Moreover, if  $f (z) =|z|^p z$ is differentiable, dose $f$ belong to the space $C^{\left\lfloor p \right\rfloor,p-\left\lfloor p \right\rfloor}(\mathbb{C})$? 
P.S. $\left\lfloor p \right\rfloor$ denotes the integer part of $p$.
 A: I assume it's real-differentiability; otherwise see DF's comment. When $p$ is an even integer, $f$ is a polynomial, hence infinitely differentiable. From now on I assume $p$ is not an even integer.
At $z\ne 0$, the function is infinitely smooth (chain rule encounters no obstacle). Observe the homogeneity: $f(tz)=t^{p+1}f(z)$ for $t>0$, and use it to conclude that the total derivative $Df$ is homogeneous of degree $(p+1)-1 = p$. 
The $k$ order derivative is homogeneous of degree $p+1-k$. 
A function that is homogeneous of degree $d$ is:


*

*differentiable at $0$ (with zero derivative) if $d>1$

*not differentiable at $0$ if $d<1$. 

*not differentiable at $0$ if $d=1$, unless it is a linear function. (The latter won't happen for us since $f$ is not a polynomial.)


All of the above are not hard to prove from the definition of derivative. 
Since we begin with homogeneity $p+1$, where $p>0$, the derivative can be taken $\lceil p \rceil $ times before its degree of homogeneity becomes $\le 1$ and the music stops. 
So, $f\in C^{\lceil p \rceil, p+1-\lceil p \rceil}$.
