How do I use Rouche's theorem here? Suppose I had the polynomial $f(z) = z^5+3z+1$ and I want to find the number of complex roots in the first quadrant. 
How would I use Rouche's theorem?  or is there a simpler way.  I was thinking of comparing it to $z$ because I checked on wolfram that it actually only has 1 root, but this is obviously not a good way to do it in general.  
EDIT: I had an idea.  Can I compare it to $z^5+1$? We would just have to show that$|f(z)-z^5-1|<|z^5+1|$? 

In order to use Rouché to find the number of roots of $f(z) = z^5 + 3 z + 1$ (counted by multiplicity) in the region inside a simple closed contour $C$, you need to find $g(z)$ such that you know the answer for $g$ and $|f(z) - g(z)|$ is relatively small (specifically $ < |f(z)| + |g(z)|$, in the best version of the theorem) on $C$.  
Well, since we can't get the whole first quadrant let's take a sector of it: take $C$ to go on the real axis from $0$ to $R$, then the circular arc $|z|=R$ in the first quadrant from $R$ to $Ri$, then back to $0$ on the positive imaginary axis.  I'll take $g(z) = z^5 + 1$: we know its roots, of which one is inside $C$ if $R > 1$.
 Now on the arc $|z|=R$, 
$$|f(z) - g(z)| = |3 z| = 3 R  < R^5 - 1 \le |g(z)| \le |f(z)| + |g(z)|$$ 
if $R$ is large enough ($R \ge 2$ certainly suffices).
On the positive real axis, 
$$|f(z) - g(z)| = 3 z < z^5 + 3 z + 1 = |f(z)| \le |f(z)| + |g(z)|$$
On the positive imaginary axis, writing $z = it$ with $t > 0$ we have
$f(z) = i(t^5 + 3 t) + 1$ so
$$|f(z) - g(z)| = 3 t < \sqrt{(t^5 + 3t)^2 + 1} = |f(z)| \le |f(z)| + |g(z)|$$
So $f$ and $g$ have the same number of roots inside $C$, namely $1$.  Taking $R \to +\infty$, we conclude that $f$ has exactly one root in the first quadrant.
