Let $v$ be the vector field on $\mathbb R^2$ defined by $v(x,y)=(x,y)$. Show that the family of diffeomorphisms $h_t:\mathbb R^2\to \mathbb R^2$ defined by $h_t(z)=e^tz$ is the flow corresponding to $v$. That is, if we fix any $z$, then the curve $t\mapsto h_t(z)$ is always tangent to $v$; its tangent vector at any time $t$ equals $v(h_t(z))$. Draw a picture of $v$ and its flow curves. Compare $\operatorname{ind}_0(v)$ with $L_0(h_t)$.
(Note that the original text defines $h_t(z)=tz$, but I used this errata list.)
The first part is "obvious" from the picture. Formally, if $z$ is fixed, then the tangent vector of $t\mapsto e^tz$ at $t$ is the derivative of this map at $t$, which is again $h_t(z)$. But $v$ is the "identity" -- $v(z)=z$, so the tangent vector, which is $\frac{d}{dt}h_t(z)=h_t(z)$, is also $v(h_t(z))$. But I have a problem with showing that the curve $t\mapsto h_t(z)$ is always tangent to $v$. This means that $\frac{d}{dt}h_t(z)=v(z)$. But this isn't true.
For the second part, $\operatorname{ind}_0(v)=1$ since the corresponding map is the identity. To compute $L_0(h_t)$, consider the map $z\mapsto \frac{h_t(z)}{|h_t(z)|}$ and compute its degree, which is $L_0(h_t)$. It seems this map is closely related to the map in the definition of the index of a (local) vector field (is it the same map?) and it seems the degree must be also 1. But how to show this rigorously?