complex-analysisroots
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?
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.
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: $t=1$
Best Answer
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.