Number of roots in the first quadrant I want to find how many roots of the equation $z^4+z^3+1=0$ lies in the first quadrant.
Using Rouche's Theorem how to find ?
 A: All the roots of the given polynomial lie inside $|z|\leq 2$: if we assume $|z|>2$ and $z^4=-(z^3+1)$, we get $2<|z|=|z^{-3}+1|<\frac{9}{8}$ which is a contradiction. Additionally it is simple to check that the roots are simple and there are no purely real or purely imaginary roots, so the number of roots in the first quadrant is given by
$$ \frac{1}{2\pi i}\oint_{\partial Q}\frac{4z^3+3z^2}{z^4+z^3+1}\,dz $$
where $Q$ is the square with vertices in $(0,0),(2,0),(2,2),(0,2)$. It is enough to approximate the (imaginary parts of the) four integrals associated to the sides of $Q$ with decent accuracy (for instance, through https://en.wikipedia.org/wiki/Newton-Cotes_formulas) to discover that there is exactly one root per quadrant.
A: Look at the family of polynomials $z^4+tz^3+1$. For $t=0$ we know the solutions $z=\sqrt{\frac12}(\pm1\pm i)$ which has one root per quadrant. We additionally know that the set of roots is continuous in the coefficients of the polynomial.
Now if changing $t$ from $0$ to $1$ were to change the number of roots in the first quadrant, one of the other roots would have to pass the positive $x$ or $y$ axis. However, on the positive $x$ axis the real part $1+x^3+x^4$ and on the $y$ axis the real part $1+y^4$ are never zero. Thus $$|z^4+tz^3+1|\ge |z^4+1|-t|z|^3>0$$ on the boundary of the first quadrant for $t\in [0,1]$. There is no change in the number of roots over this homotopy.

roots of $z^4+tz^3+1$ for red: $t=0$ over blue to green: $t=1$
A: For completeness, here I present a more standard solution to this problem:
Let $p(z):=z^4+z^3+1$. Consider the following path

Since $p(x)>0$ and $p(ix)=x^4+1-ix^3\neq 0$ for $x\geq 0$, we deduce that $p$ has no roots on $\gamma_R$. Therefore, we can apply the Argument Principle to obtain that the number of roots of $p$ in the first cuadrant is the winding number of $0$ around $p\circ \gamma_R$, for sufficiently big $R$.
Besides, for big $R$ the argument variation of $p\circ \text{quarter circle}$ is approximately $\Delta \arg z^4=4\frac{\pi}{2}=2\pi$. Therefore, this fact combined with $p(x)>0$, $p(ix)\in 4\text{th quadrant}$ tells us that $p$ has exactly one root in the first quadrant.
