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.


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.

  • $\begingroup$ 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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