Conditions that a function is analytic in the complex plane of its independent variable? I have no a mathematic undergraduate background, so I am very sorry if this question is too naive.
Consider a simple example: $f(x)=\vert x \vert^3$ and $g(x)=x^3$ where $x\in \mathbb{C}$. Why $f(x)$ is not analytic in the complex $x$ plane and $g(x)$ is a analytic function in the extire complex plane of $x$? or what is the conditions that a function is analytic in the complex plane of its independent variable?
Please explain as detail as possible but please not use too much jargon. Thank you very much.
 A: Start with the definition of analytic.  The function of a complex variable $z$ is analytic at $z \in \mathbb C$ if it is differentiable at $z$, which means 
$$\begin{align} \frac{f(z+h) - f(z)}{h} \tag 1 \end{align}$$ 
has a unique limit as $\lvert h \rvert \to 0$, denoted $f'(z)$.  The limit has to exist regardless of how and in which direction $h$ approaches zero.
This is a strong requirement and requires $f(z)$ to satisfy the Cauchy Riemann equations.  
These are obtained as follows:  write $z=x+iy$ and $f(z) = u(x,y)+iv(x,y)$ and consider the complex derivative when $h = \delta x$ and $h = i\delta y$ for real $\delta x,\delta y$.  If $f$ is required to be analytic at $z$ then both are must be the same, so we obtain,
$$\frac{\partial f}{\partial x} = f'(z) = -i \frac{\partial f}{\partial y}.$$
Now write this in terms of $u,v$ to obtain the Cauchy-Riemann equations,
$$ \begin{align}
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \quad
\frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}. \tag 2 
\end{align}$$
When applied to $\lvert z \rvert^3$ these break down.  We have $u(x,y) = (x^2+y^2)^{3/2} $ and $v(x,y) = 0$.  It is not difficult to see then that $(2)$ will only be satisfied by exception, when $x = y = 0$.  Thus $\lvert z \rvert^3$ cannot be analytic except at the single point $z = 0$.
I hope this is useful.
