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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.

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1 Answer 1

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I don't know what you mean by "does it represent the domain range?".

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 $|\zeta_1| = 1$.

The product of two functions is analytic in a region where both of the functions are analytic.

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