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 ?
Number of roots in the first quadrant
complex-analysis
Related Solutions
First apply Rouche's theorem to find the number of zeros inside the unit disk. Here you need a function which dominates on the boundary of the unit disk and whose zeros inside you know. Note that on the boundary (when $|z|=1$) $-8z^3$ is quite large compared to the other terms of your polynomial $z^9+z^5-8z^3+2z+1$. You should get that there are 3 roots in the disk of radius 1.
Next apply Rouche's theorem on the disk of radius 2. Here you need a function which dominates when $|z|=2$ and whose zeros inside the disk are known. Note that $z^9$ is quite large when $|z|=2$. You should find there are 9 zeros in the disk of radius 2.
Now for the final answer, subtract the number of zeros in the disk of radius 1 from the number of zeros in the disk of radius 2.
Let $\Gamma _1: z=x$ $(0\le x\le R)$, $C: z=Re^{i\theta } $ $( 0\le \theta \le \pi/2)$
and $\Gamma _2: z=i(R-y)$ $(
0\le y\le R)$ as in Fig.1.
We note
several facts:
(1) For $z=x$ real, $f(x)=x^4+2x^3+3x^2-x+2>0 $ $(\, -\infty< x<\infty)$,
(2) For sufficiently large $z$, $\arg f(z)=\arg z^4(1+o(1/z))=\arg z^4+\arg (1+o(1/z))
=4\arg z+\varepsilon $,
(3) For $z=iy$, where $y$ is real positive, $\Im f(iy)=-2y^3-y<0$.
From these facts, $f(\Gamma _1), f(C)$ and $f(\Gamma _2)$ are like as in Fig.2 if $R$
is suffiently large. They go around the origin once, which means that the number of
zeros of $f(z)$ inside the contour $\Gamma _1 \cup C
\cup \Gamma _2$ is $1$ by the argument principle.
Same arguments can be applied for quadrant II and the number of zeros of $f(z)$ inside
quadrant II is $1$.
If $\alpha $ is a root of $f(z)=0$, then $\overline{\alpha}$ is also
a root of it. Thus the number of zeros of $f(z)$ is $1$ in each quadrant.
Best Answer
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: $t=1$