# Clarifying analyticity in complex functions and domain ranges

I'm confused about the concept of analyticity in complex functions. How can I determine the domain of analyticity for a function?

For example: 1.

$${f_1}({\zeta _1}) = \frac{{\zeta _1^2 + m}}{{1 - m\zeta _1^2}},\quad \left| {{\zeta _1}} \right| \le 1$$

Since $$(0 \le m \le 1)$$, $$f_1$$ is an analytic function within the unit circle.

$$g_1(\zeta_1) = -\zeta_1 - 2\zeta_1^2 - 3\zeta_1^3 \cdots, \quad |\zeta_1| \le 1$$

$$g_1(\zeta_1)$$ is an analytic function within the unit circle as well.

$$f_1(\zeta_1) \cdot g_1(\zeta_1)$$ This function is also analytic within the unit circle.

2:

$$f_0(\zeta_0) = \frac{1 + m\zeta_0^2}{\zeta_0^2 - m}, \quad |\zeta_0| \ge 1$$

Due to $$(0 \le m \le 1)$$, $$f_0$$ is an analytic function outside the unit circle, including the point at infinity. $$g_0(\zeta_0) = -\frac{1}{\zeta_0} - 2\frac{1}{\zeta_0^2} - 3\frac{1}{\zeta_0^3} \cdots$$ $$g_0(\zeta_0)$$ is an analytic function outside the unit circle, including the point at infinity as well.

Where a function is analytic, does it represent the domain of definition? Why does it relate to $$m$$?

This question has been bothering me for a while. I look forward to your help.

A rational function such as $$f_1$$ is analytic everywhere in $$\mathbb C$$ except at the zeros of its denominator. In this case the zeros are $$\zeta_1 = \pm 1/\sqrt{m}$$; since $$0 \le m \le 1$$, those zeros have absolute value $$\ge 1$$, i.e. they are not inside the unit circle, so $$f_1$$ is analytic inside the unit circle. Not just inside the unit circle, but apparently here we are only interested in the region inside the unit circle.
The sum of a power series such as $$g_1$$ is analytic inside its circle of convergence. In this case the radius of convergence is $$1$$, so $$g_1$$ is analytic inside the unit circle. BTW, $$|\zeta_1| \le 1$$ is incorrect here, it should be $$|\zeta_1| < 1$$: this series diverges when $$|\zeta_1| = 1$$.