Let $v(x,y)=(-y,x)$. Show that the flow transformation are the linear rotation maps $h_t: \mathbb R^2\to \mathbb R^2$ with matrix \begin{bmatrix}\cos t&-\sin t\\\sin t&\cos t\end{bmatrix} Draw $v$ and its flow curves. Also compare $\operatorname {ind}_0(v)$ with $L_0(h_t)$.
Please note that (according to this question, which is the exercise in Guillemin and Pollack preceding the one above), the flow in this context is defined in this way: $h_t$ is the flow if for $z$ fixed, the curve $t\mapsto h_t(z)$ is always tangent to $v$. According to p.135 of the book (Guillemin and Pollack) this means that $\frac{d}{dt}h_t(z)=v(z)$ (actually there the derivative is taken at $t=0$, but I believe "always" means "for all $t$").
So in the above context if $z=(x,y)$, then $h_t(x,y)=(x\cos t-y\sin t, x\sin t+y\cos t)$, and $\frac{d}{dt}h_t(x,y)=(-x\sin t-y\cos t, x\cos t-y\sin t)$. This must be equal somehow to $v(x,y)$. If $t=0$, this holds. But again, I thought "always" means "for all $t$", is this incorrect assumption? Do I only need to show that $t\mapsto h_t(z)$ is tangent to $v$ at $t=0$?
Regarding the second part, the map from the definition of $\operatorname {ind}_0(v)$ seems to be a rotation, so it is homotopic to the identity, and hence its degree is $1$, thus $\operatorname {ind}_0(v)=1$. To compute $L_0(h_t)$, I believe I need to consider the map from a small ball containing no other zeros of $v$ to $S^1$ given by $z\mapsto \frac{h_t(z)}{|h_t(z)|}$ and compute its degree. Am I correct in saying that this map eats a point, rotates it by angle $t$, and spits its projection onto $S^1$? If so, this map seems to be homotopic to the identity and also has degree 1.