The question is:
Show that $$ P(z) = z^4 + 2z^3 + 3z^2 + z +2$$ has exactly one root in each quadrant of the complex plane.
My initial thought was to use Rouche's Theorem (since that's generally what I use to find how many roots a complex polynomial has), but the more I think about it, the more I'm not sure how to make it work. Here is my attempt:
First, pick a radius for a circle that can encompass all four roots of the polynomial. For simplicity sake (in my opinion), I went with |z| = 5.
Setting $$f(z) = z^4$$ and $$g(z) = 2z^3 + 3z^2 + z + 2$$
I get |f(z)| = 625 and |g(z)| = 332, so by Rouche's Theorem we have four roots in the disc.
Now, my thought was that I could somehow seperate the quadrants by breaking up my circle into quarter-circles (like four slices of pie) and applying Rouche's Theorem again on each of these new domains. However, finding the place on the boundary where the value hits its max for some f(z) or g(z) would be messy (at best), since if I juse use |z| = 5, I'm right back where I started. There's also the issue that these zeroes may occur on the boundary, so in the real/imaginary axis, which wouldn't be what I'm trying to show. So now, I'm just stuck, so if anyone can see how to tackle this, it would be greatly appreciated.
Best Answer
First dispose of real roots: e.g. $P(z) = z^2 (z+1)^2 + 2 (z+1/4)^2 + 15/8$.
Let's look at what happens to $P(z)$ as $z$ goes around a contour around part of the first quadrant. As $z$ goes from $0$ to some large positive $R$ on the real axis, $P(z)$ increases from $2$ to $P(R) >> 0$. Then go on the quarter-arc of the circle $|z| = R$ from $R$ to $iR$: $P(z)$ goes almost in a circle, ending at $P(iR)$ which is in the fourth quadrant. Now come back in to the origin on the imaginary axis. Note that $\text{Re}(P(it)) = t^4 - 3 t^2 + 2 = 0$ at $t = 1$ and $t=\sqrt{2}$, while $\text{Im}(P(it)) = - 2 t^3 + t = 0$ at $t=0$ and $t = \sqrt{2}/2$. So you hit the negative imaginary axis at $t=\sqrt{2}$ and again at $t=1$, then the positive real axis at $t=\sqrt{2}/2$ and $t=0$, but not the negative real or positive imaginary axis. Thus as $z$ goes around this contour, the winding number of $P(z)$ around $0$ is $1$, indicating that there is exactly one zero of $P(z)$ inside the contour.
Here's a plot of the case $R=1.6$: