# Laurent's Theorem - definition of a path in an annular domain

Theorem on p. 197 of the book mentioned below:

Supposed that a function $f$ is analytic throughout an annular domain $R_1<\lvert z-z_0\rvert< R_2$, cenetered at $z_0$, and let $C$ denote any positively oriented simple closed ontour around $z_0$ and lying in that domain. Then, at each point in the domain, $f(z)$ has the [Laurent] series representation.

In Complex Variables and Applications by Brown and Churchill (Ninth Edition), on page 201, the case of $z_0 \ne 0$ in the Laurent's Theorem is proved (i.e., that Laurent series exists when the annular domain is not centred at $z=0$).

In this proof, the authors consider the annulus $R_1 < \lvert z-z_0 \rvert < R_2$, where $z_0 \ne 0$. They define $g(z) := f(z+z_0)$ and so $g(z)$ is analytic on $R_1 < \lvert z \rvert < R_2$.

Next they say that the simple closed contour $C$ in the statement of the theorem has some parametric representation $z=z(t)$ $(a\le t \le b)$, where $$R_1 < \lvert z(t)-z_0 \rvert < R_2\text{ (*)}$$ for $t\in[a,b]$. OK, this is clear so far.

But here's where comes a part I don't quite understand:

... if $\Gamma$ denotes the path $z=z(t)-z_0$ $(a\le t \le b)$, $\Gamma$ is not only a simple closed contour but, in view of inequality (*), it lies in the domain $R_1<\lvert z\rvert < R_2$.

How is inequality (*) related to $\Gamma$ lying in $R_1<\lvert z\rvert < R_2$?

Another thing I don't understand is that they use the parametrization $z=z(t)-z_0$. But this would not be centred about $z_0$, would it? We can represent a complex number $z$ as a vector in $\mathbb{R}^2$. And, for example, a circle centred at $(1,1)$ in $\mathbb{R^2}$ would be parametrized as $(\cos(t), sin(t)) + (1,1)$, not as $(\cos(t), sin(t)) - (1,1)$. So why are they using the minus sign?

We have a curve $C=z(t)$ centered at $z_0$. We stipulate that $C$ lies in the domain of the function, namely the annulus $R_1<|z-z_0|<R_2$. Hence, we have inequality (*), which states
$$R_1<|z(t)-z_0|<R_2$$
Now, we define $\Gamma$ to be a new curve $z=z(t)-z_0$, which is distinct from $C$ (a translation in $\mathbb{C}$, to be precise). Therefore, there is no concern about the minus sign, because $\Gamma$ is not supposed to be centered at $z_0$; $C$ is.
For example, if $C$ was a circle of radius $1$, we could paramaterize it by $z(t)=e^{it}+z_0$. In this case, the curve $\Gamma$ would have parameterization $e^{it}+z_0-z_0=e^{it}$, i.e. it would be centered at $0$.
Now, recall that we defined $z=z(t)-z_0$, so it follows immediately from (*) that $R_1<|z|<R_2$. This simply tells you that the curve $\Gamma$ lies in the annulus $R_1<|z|<R_2$, which is centered at $0$.