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.

  • 4
    $\begingroup$ If you can show that there is a root in the upper right quadrant, then there is automatically one in the lower right, and likewise for the upper and lower left quadrants, due to the fact that the complex conjugate of a root is also a root. $\endgroup$ – Emily Jul 24 '12 at 22:14
  • $\begingroup$ That certainly makes the problem a little bit easier, but I'm still stuck at finding the root in the upper right quadrant. At least, using my method (and at the moment I'm unsure of a different approach). $\endgroup$ – Bradley Jul 24 '12 at 22:40
  • $\begingroup$ Here is the maple solution! .1097234460+.8436472749*I, -1.109723446+1.237652874*I, -1.109723446-1.237652874*I, .1097234460-.8436472749*I $\endgroup$ – PAD Jul 25 '12 at 6:20

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 $R=1.6$:

enter image description here

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    $\begingroup$ I get most of your argument, but I get a little lost when you bring up the winding number. In this case, how are you getting that the winding number is 1? $\endgroup$ – Bradley Jul 25 '12 at 18:46
  • $\begingroup$ Geometrically. Look at the picture. The curve (which is going counterclockwise) winds once around the origin. $\endgroup$ – Robert Israel Jul 25 '12 at 19:54
  • $\begingroup$ Oh wow. Yeah, I'm with you now. Thanks! $\endgroup$ – Bradley Jul 25 '12 at 20:12

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