why $f(z) = z^{(3/2)}$ does not have derivative at z = 0 in complex plane. it seems that the $f'(z) = z^{(1/2)}$ means that this function has derivative for every complex value. But why $f(z) = z^{(3/2)}$ does not have derivative at z = 0
 A: The function $f(z)=z^{3/2}$ is not single-valued in the entire plane.  
We can see this easily by noting that $z=re^{i\theta}=re^{i(\theta+2n\pi)}$.
However, $z^{3/2}=(re^{i(\theta+2n\pi)})^{3/2}=(-1)^n(re^{i\theta})^{3/2}$.
This means that for each point in the $z$-plane, there are two points mapped by $z^{3/2}$.
Now, if one deletes ("cuts") from the plane a contour that begins at the origin and extends to the point at infinity, then $f$ is single-valued and analytic in this "cut plane."  
For example, one can choose to cut the plane along the negative real axis, from $(0,0)$ to $(-\infty,0)$.  Then, if we define the argument $\theta$ of $z$ so that $-\pi<\theta\le \pi$, then $f(z)=z^{3/2}$ is single valued and analytic in the plane less the non-positive real numbers.
In the cut plane, we can therefore assert that the integral of $z^{1/2}$ along any contour $C$ that (i) does not cross the negative real axis (or the origin) and (ii) begins at $z_1$ and terminates  $z_2$ is given by
$$\int_Cz^{1/2}\,dz=\frac23\left(z_2^{3/2}-z_1^{3/2}\right)$$
and that the derivative of $z^{3/2}$ is $z^{1/2}$ in the cut plane.
