How many zeros does the function $f(z)=z^3+z^2+4z+1$ have in the first quadrant? 
How many zeros does the function $f(z)=z^3+z^2+4z+1$ have in the first quadrant?

Using Rouche's theorem with the function $h(z)=1,g(z)=z^3+z^2+4z$, I've been able to show that all the roots are in $B_3(0)$.
Is it possible to show that they are not in $B_3(0)\cap\{Re(z),Im(z)\ge0\}$ using Rouche's theorem?
 A: Since $f'(z)>0$ for all real $z$ the function has exactly one real root. Calculating the value of the function at $z=0$ and $z=-1$ one concludes that the real root satisfies the inequality $-1<z_1<0$. Further since:
$$
z_1+z_2+z_3=-1
$$
one finds that $\Re(z_2)=\Re(z_3)<0$. Therefore the polynomial has no roots in the first quadrant.
A: The answer by @user is clearly how to tackle this specific question.
But if you are interested in a Rouche/Argument Principle way to tackle such problems let me say how I'd do this one.
$f(z)$ a polynomial, so by the Argument Principle the number of zeros in a region inside a simple contour is $2\pi$ times the increase in the argument of $f(z)$ as we go round the countour.
So consider a contour consisting of (i) the straight line from $0$ to $R$, (ii) a circular arc of radius $R$ from $R$ to $R i$, (iii) the straight line from $Ri$ back to the origin.
Along (i) $f(z)$ is real, and so its argument remains $0$.
Along (ii) we have
$$
\arg f(z)=\arg z^3 +\arg (1+\frac{1}{z}+4\frac{1}{z^2}+\frac{1}{z^3})
$$
which simplifies when $z=Re^{i\theta}$ to
$$
\arg f(z)=3\theta +\arg (1+O(\frac{1}{R})).
$$
Hence as we run round the arc the argument increases by $\frac{3\pi}{2}$.
Along (iii) we have $z=iy$ and so
$$
f(z)=-iy^3 -y^2 +4iy +1
$$
whose argument is
$$
\tan^{-1}Y, \text{  where  }Y=\frac{y(y-2)(y+2)}{(y-1)(y+1)}.
$$
We now need to make a little sketch of $Y$ as $y$ varies -easier on a piece of paper than it is here - and what we see is that at $+\infty$ it starts at $+\infty$, it decrease as $y$ decreases, cutting the axis at $y=2$, runs off to $-\infty$ at $y=1+$, comes back at $+\infty$ at $y=1-$, then decreases to $0$ at $y=0$.
Now we have to account for what the argument of $Y$ does. We got at the end of (ii) to $\frac{3\pi}{2}$, so we start there. When we get to $y=2$ the argument is now $\pi$, and by the time we get to $y=1+$ we are down to $\frac{\pi}{2}$; now we hop on to the next branch of tangent, and as we pass from $y=1-$ to $y=0$ the argument runs down to $0$.
Summing up, as we travel round the whole contour the argument has not changed at all. Hence there are no zeroes of $f(z)$ in the first quadrant.
