How many zeros does $z^{4}+z^{3}+4z^{2}+2z+3$ have in the first quadrant? Let $f(z) = z^{4}+z^{3}+4z^{2}+2z+3$. I know that $f$ has no real roots and no purely imaginary roots. 
The number of zeros of $f(z)$ in the first quadrant is $\frac{1}{2\pi i}\int_{C}\frac{f'(z)}{f(z)}\, dz$ by the Argument Principle where $C = C_{1} + C_{2} + C_{3}$ with $C_{1}$ the line from origin to the point $(R, 0)$ on the $x$-axis, $C_{2}$ is the line $Re^{i\theta}$ where $0 \leq \theta \leq \pi/2$, and $C_{3}$ is the line segment from the point $(0, R)$ to the origin.
We have
$$\frac{1}{2\pi i}\int_{C_{2}}\frac{f'(z)}{f(z)}\, dz = \frac{1}{2\pi}\int_{0}^{\pi/2}4 + O(R^{-1})\, d\theta \rightarrow 1$$ as $R \rightarrow \infty$.
I am having an issue computing the integral for $C_{1}$ and $C_{3}$. For $C_{1}$, I want to compute
$$\lim_{R \rightarrow \infty} \frac{1}{2\pi i}\int_{0}^{R}\frac{4x^{3} + 3x^{2} + 8x + 2}{x^{4} + x^{3} + 4x^{2} + 2x + 3}\, dx$$ but doesn't this evaluate to $\infty$ since the inside is $O(1/x)$?

Here's an alternative idea, which can be applied to any quartic.
Let $z=x+iy$, expand the powers of $z$, and separate the real and imaginary parts to get
$$x^4+x^3+4x^2+2x+3-(6x^2+3x+4)y^2-y^4=0$$
and
$$(4x^3+3x^2+8x+2)y-(4x+1)y^3=0$$
Since we looking for solutions with $x,y\gt0$, we can rewrite the second equation as
$$y^2={4x^3+3x^2+8x+2\over4x+1}$$
and substitute into the first, obtaining (with a multiplication by $-1$)
$$(4x^3+3x^2+8x+2)^2+(4x+1)(4x^3+3x^2+8x+2)(6x^2+3x+4)-(4x+1)^2(x^4+x^3+4x^2+2x+3)=0$$
Technically one should expand this out, but it seems fairly clear that the coefficients from the first two, positive portions will more than cancel the negative coefficients from the third, leaving a sextic with all positive coefficients and hence no roots with $x\gt0$.
For a general quartic with real coefficients, you can write $y^2$ as a cubic divided by a linear expression in $x$, producing a sextic, which in general may have some positive roots.  It may sound as if you could wind up with $6$ real values for $x$, each with two possible values for $y$, giving $12$ roots in all to a quartic.  Obviously something's gotta give.  Either the sextic has fewer than $4$ real roots, or some of its real roots produce a negative value for $y^2$ (or both).
