I'm studying Lee's introduction to smooth manifolds, in chapter 9 he introduces integral curves and vector flows, I'm using the following definitions.
Definition: Let M be a manifold a flow domain for M is an open subset D $\subset \mathbb{R} \times M $ such that $\forall p \in M$ $D^{(p)}= \{ t\in \mathbb{R} | (t,p) \in D\} \ $ is an open interval containing $0$.
Definition: Let M be a manifold a flow on M is a continuous map $\theta : D \to M$ where D is a flow domain such that $\theta$ satisfies the following group laws
$\bullet \forall p \in M \ \ \theta(0, p) = p$
$\bullet \forall s\in D^{(p)}$ and $t \in D^{(\theta(s,p))}$ such that $s+t \in D^{(p)}$ $$ \theta(t, \theta(s,p))= \theta (s+t, p)$$
We also defined $\theta_t(p) = \theta^{(p)}(t) = \theta(t,p)$ and $M_t=\{p\in M| (t,p) \in D\}$
I'm also familiar with the fundamental theorem of flows
I'm asked to find the flow of the vector field $V(z)=\frac{1}{\overline{z}}$ in the manifold $\mathbb{C}-0$ and the image of $\theta_t(A)$ for $t>0$ and A is the annulus $1<|z|<2$.
I realized that I needed to find all integral curves starting at arbitrary points $(x_0, y_0) $ in M and for that I solved the system of differential equations \begin{cases} x'(t) = \frac{x}{x^2+y^2}\\ y'(t)= \frac{y}{x^2+y^2}\\ x(0)= x_0\\ y(0)=y_0 \end{cases} where $x_0, y_0$ can't be both $0$, I solved the system using substitution and separable equations and found the most general solution to be $$\theta^{(x_0,y_0)}(t) = (\sqrt{\frac{2t}{(y_0/x_0)^2 + 1}+ x_0^2}) \ (\text{sgn}(x_0), \frac{y_0}{|x_0|})$$ for $t>-(1/2)(x_0^2+y_0^2)$ when $x_0 \neq 0$
and $$\theta^{(0,y_0)}(t) = (\sqrt{2t + y_0^2})(0, \text{sgn}(y_0))$$
for $t>\frac{-y_0^2}{2}$
All this already gives me the flow of the vector field because of the uniqueness of the maximal integral curves and I know how the domain flow looks like $\mathbb{R}^3$ minus a paraboloid and the $z$ axis.
My question is, are the integral curves that I found okey ? and how do I find $\theta_t(A)$ for $t>0$? I'm stuck there and I don't know how to proceed.