Let $p(z) = z^6 + 9z^4+z^3+2z+4$
- find then number of roots in each quadrant of the complex plane
- find in which quadrant exists a root which is inside the unit circle
using the Argument priniciple I was able to determine that there are 2 solutions in the first quadrant so there are 2 solutions in the 4 quadrant
because there are 6 solution to this polynomial, there are 2 solutions in the left half plane.
so there are 2 options, they are either both on the real axis, or 1 on the 2 quadrant and one on the 3 quadrant.
I know the answer is 1 and 1, But dont know how to show it?
for the second part, using Rouché's theorem its easy to see there are 4 roots inside the unit circle, but how do I show on which quadrant they are?