how to think of order of infinity of rational functions I am just reading some lecture notes on complex analysis. 
I am trying to intuitively understand what is happening with this order of infinity stuff? Can someone explain ?
 A: The idea is that the order describes how $R(z)$ vanishes or blows up as $z\to z_0.$ Basically, the rational function can have asymptotic behavior of the following three kinds: $$ R(z)\sim C\\R(z)\sim C(z-z_0)^p \\ R(z) \sim \frac{C}{(z-z_0)^{p}}:$$ either it goes to a constant (in which case the order at $z_0$ is zero) or it vanishes as $(z-z_0)^p$ for some positive integer $p$ (in which case it has order $p$) or it can blow up (have a pole) as $(z-z_0)^{-p}$ for some positive integer $p$, in which case it has order $-p.$
Now we want to extend that to the behavior as $z\to\infty.$ The asymptotics here can also look like one of three things $$R(z) \sim C \\ R(z)\sim Cz^p \\R(z)\sim \frac{C}{z^p}.$$ For instance $$ R(z) = \frac{z^2-iz+2}{3z^2+2z+3}\sim \frac{1}{3}\\R(z) = \frac{z^4+1}{iz^2-4\pi} \sim \frac{z^2}{i}\\ R(z) = \frac{3z-4}{z^5} \sim \frac{3}{z^4}.$$ In the first case we say the order is zero again.  Since we're at infinity now, in the second case we're blowing up and should want to say that $R$ has a pole of order two at $z=\infty.$ So to match the previous convention, the order is $-2.$ Similarly in the third case $R$ vanishes as the fourth power at infinity, so the order is $+4.$
As for the definition you were given, it should be clear that as $w\to 0$ we have respectively to the previous example: $$ R(1/w)\sim_{w\to 0} \frac{1}{3}\\R(1/w)\sim \frac{1}{iw^2} \\ R(1/w)\sim 3w^4$$ so according to the definition in your notes we get $Ord_\infty=0,-2,4$ as expected.
