What is the Taylor series expansion of $z^{1/2}$ about origin. Consider $z^{1/2}={re^{i\theta}}^{1/2}=r^{1/2}e^{i\theta/2}$
Now seprate into real and imaginary parts we have
$U=r^{1/2}\cos(\theta/2)$
$V=r^{1/2}\sin(\theta/2)$
Now $rU_r=1/2r^{1/2}\cos(\theta/2)$ and $V_{\theta}=1/2r^{1/2}\cos(\theta/2)$ So it satisfies CR equations and thus the function is analytic.
Now my question is what is the Taylor series expansion of $z^{1/2}$about $z=0$
Moreover what is the vavlue of ${(0+i0)}^{1/2}$
 A: When you write $z$ in a polar form in an expression, you are not writing something in terms of $z$, but in terms of $\log z$.
You have said $z^{1/2} = r^{1/2} \exp(i\theta/2)$ but this is not a well-defined function of $z$ for example if $z= -1$ then according to the choice of $\theta$ you can obtain $(-1)^{1/2} = 1$ or $(-1)^{1/2} = -1$.
So this expression is only a function of $r$ and $\theta$ (or equivalently, of $\log z = \log r + i \theta$).
Now what you have written in terms of $l = \log z$, is that "the square root of $\exp l$" should be $ \exp (l/2)$.
Now $l \mapsto \exp(l/2)$ is a nice holomorphic function and everything, so is $l \mapsto \exp l$ and its derivative never vanishes, so by the implicit function theorem, if you pick some $z$ in the image of $\exp$ there is a small open subset containing $z$ where you can have a well-defined map $z = \exp l \mapsto l \mapsto \exp(l/2) = \sqrt z$ which is holomorphic.
But of course, $0$ is not in the image of $\exp$ (and of course $\log z$ and $\sqrt z$ can't be well defined on a neighbourhood of $z=0$). So whenever you use a polar form, the fact that it looks holomorphic, doesn't mean that it is holomorphic at $z=0$.
