# Thread: Quadrants, Zeroes

1. ## Quadrants, Zeroes

Let $\displaystyle p(z)=z^6+9z^4+z^3+2z+4$.

Determine which quadrants contain the two zeros of
$\displaystyle p(z)$ that lie outside the unit circle.

We have learned the argument principle in this section. However, I
don't see how to find the quadrants that the zeroes lie in. I wanted
to use De Moivre's formula; however, I don't think this problem is
just that easy.

2. Originally Posted by vampira673
Let $\displaystyle p(z)=z^6+9z^4+z^3+2z+4$.

Determine which quadrants contain the two zeros of
$\displaystyle p(z)$ that lie outside the unit circle.

We have learned the argument principle in this section. However, I
don't see how to find the quadrants that the zeroes lie in. I wanted
to use De Moivre's formula; however, I don't think this problem is
just that easy.
I think you can do this by the argument principle as follows. If z = iy is purely imaginary then $\displaystyle p(iy) = -y^6 + 9y^4 +4 -iy(y^2-2)$. Now think about what happens to p(z) as z goes round a contour consisting of the imaginary axis from +iR to –iR (where R is a large positive real number) and then round the semicircle $\displaystyle z= Re^{i\theta}\ (-\pi/2\leqslant\theta\leqslant\pi/2)$.

As z goes down the imaginary axis, p(z) goes (almost) once round the origin anticlockwise: when z=+iR, p(z) is dominated by the term $\displaystyle z^6 = -R^6$ (which is very large and negative), with a comparatively small negative imaginary part, roughly $\displaystyle -iR^3$. As z goes down the imaginary axis, p(z) crosses the negative imaginary axis when z is approximately 3i, then it crosses the positive real axis when $\displaystyle z=\sqrt2i$. It then does a little loop in the right half-plane, crossing the positive real axis again when z=0 and $\displaystyle z = -\sqrt2i$, before crossing the positive imaginary axis when z is approximately –3i, finally going off towards $\displaystyle -R^6$ as z heads towards –iR.

Now complete the contour by sending z round the semicircle. Here, the $\displaystyle z^6$ term dominates, and p(z) essentially circles the origin three times anticlockwise. Thus over the entire contour, p(z) goes four times round the origin, and the argument principle then tells us that p(z) has four zeros in the right half-plane.

I guess you should already have shown in an earlier part of the question (using Rouché's theorem) that, inside the unit circle, p(z) has one zero in each quadrant. It then follows that the two zeros outside the unit circle lie in the right half-plane. They obviously form a complex conjugate pair, since the equation has real coefficients, so one of them is in the first quadrant, and the other one in the fourth quadrant.