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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.

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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.

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  • $\begingroup$ Thank you. Do you mean (1) should have a unique limit as $\vert h\vert \to0$ because it seems that $h\in \mathbb{C}$? $\endgroup$
    – user55777
    Jun 4 '20 at 8:59
  • $\begingroup$ Exactly so. It is not unusual to say $h \to 0 $ and $|h| \to 0 $ interchangeably. I'll amend the answer. $\endgroup$
    – WA Don
    Jun 4 '20 at 9:29

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