How many zeros does $z^{4}+z^{3}+4z^{2}+2z+3$ have in the first quadrant?

Try to combine $C_1$ and $C_3$ before taking the limits

There is another technique that is seldom shown to students in its full strength: the Rouche theorem, which says that if $f_t(z)$ ($0\le t\le 1$) is a continuous family of analytic functions and $G$ is a nice domain, then if no $f_t$ nas any zeroes on $\partial G$, the functions $f_0$ and $f_1$ have the same number of zeroes in $G$.

Here we can take $G$ to be sufficiently large quarter-disk. The quarter circle is not a concern as long as we keep the term $z^4$ intact. The positive radius is not a problem either if we don't kill $3$ or introduce negative coefficients in our homotopy. Thus, the imaginary radius is the main part of the game.

We need some obvious reason at least for the initial polynomial not to vanish there and we can find one: just notice that odd powers are imaginary and even ones real. So, the only chance for zero is to have both $z^3+2z=0$ and $z^4+4z^2+3=0$ simultaneously. However, if $z^2=2$ or $0$, the second expression is an odd integer.

This strongly suggests keeping the roots of the first expression and the second expression intact, i.e., to take the homotopy $f_t(z)=z^4+4z^2+3+t(z^3+2z)$.

$f_1$ is what we need and $f_0(z)=z^4+4z^2+3$ is a biquadratic polinomial, whose roots are trivial to find since it factors as $(z^2+1)(z^2+3)$. Unfortunately, it has 4 roots right on the imaginary axis, so it cannot be used directly in the formal way.

Fortunately, we've learned enough in the process to make an effective twist of the argument and finish. Our polynomial is $(z^2+1)(z^2+3)+z(z^2+2)$ and we can do any homotopy $(z^2+a(t))(z^2+c(t))+q(t)z(z^2+b(t))$ that keeps $0<a(t)<b(t)<c(t)$ and $q(t)>0$ all the way.

Now just note that $g(z)=z^4+6z^2+1+4z(z^2+1)$ has the same structure because $z^4+6z^2+1$ has two different real roots with product $1$, which both must be negative, so it factors as $(z^2+a(t))(z^2+c(t))$ with $a(t)<b(t)=1<c(t)$. Since we can continuously move any triple of positive numbers into any other such triple ordered the same way without violating the order on the way (the trivial linear homotopy will already do the trick), we see that our initial polynomial has as many roots in the first quadrant as $g(z)$. But $g(z)=(z+1)^4$, so we don't have to think too hard about its number of roots in the first quadrant.

As a byproduct, we've got a simple "human accessible" test for figuring out when a 4-th degree monic polynomial with real coefficients has all its roots in the left half-plane. The coefficients must, clearly, be positive (even in each quadratic factor) and after that all depends on the order of $a,b,c$ in the above notation.

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).