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

  1. find then number of roots in each quadrant of the complex plane
  2. 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?



Concerning the first part:

You want to prove that there is no real root to the polynomial $p(z)=z^6+9z^4+z^3+2z+4$.

Now, the idea is to group terms to prove that $p(z)>0$ for all real $z$.

First of all, you can get away that $z^3$ term using $z^6+z^3+\frac{1}{4} \ge 0$ which is equivalent to $(z^3+\frac{1}{2})^2 \ge 0$ which holds for all real $z$.

Now, it suffices to prove $9z^4+2z+4 \ge 0$.

For $z>0$ this is clearly true. Assume $z<0$. Then we want to prove that $9x^4+4 \ge 2x$ holds for all $x>0$. But by the AM-GM inequality you have $9x^4+3=4 \cdot \frac{9x^4+1+1+1}{4} \ge 4 \cdot \sqrt[4]{9x^4 \cdot 1 \cdot 1 \cdot 1}=4x \cdot \sqrt{3}>4x>2x$. So $9x^4+4 >2x+1>2x$ holds for all positive $x$.

Hence $p(z)$ has no real roots i.e. there must be one root in each the second and the third quadrant.

  • $\begingroup$ It is true that $p$ has no real root, but these lines do not prove anything about the location (i.e. the quadrants in which they lie) of the roots of $p$ inside/outside the unit disk. $\endgroup$ – Jack D'Aurizio Aug 30 '15 at 17:40
  • $\begingroup$ @JackD'Aurizio: Sure. This is why I started my answer with "Concerning the first part" $\endgroup$ – Tintarn Aug 30 '15 at 17:51

$p$ is trivially increasing on $\mathbb{R}^+$, hence its real zeros are negative. However, $x\leq -1$ implies $p(x)\geq 11$, hence the real roots lie in $(-1,0)$. Since: $$ p(z)=(z^6+9z^4+1)+(z+1)(z^2-z+3)$$ there are no real roots at all. Since $1+1+2+4<9$, by Rouché's theorem there are exactly four roots of $p(z)$ inside the unit disk, that obviously come in conjugated couples. The number of roots in each set $\text{Re}(z)>0,\text{Im}(z)>0,\text{Re}(z)<0,\text{Im}(z)<0$ can be computed by applying a Cayley transform, then Rouché's theorem. For instance, if $w$ is a root of $p$ with positive real part, then $\frac{w-1}{w+1}$ lies inside the unit disk and $\frac{w-1}{w+1}$ is a root of:

$$p\left(\frac{1+z}{1-z}\right) = \frac{17 - 8 z + 73 z^2 - 96 z^3 + 59 z^4 + 8 z^5 + 11 z^6}{(1-z)^6}$$ hence there are four roots with positive real part and two roots with negative real part. Hence there are two cases: if the two roots outside the unit circle lie in $\text{Re}(z)<0$, all the roots inside the unit circle lie in $\text{Re}(z)>0$; otherwise the two roots outside the unit circle have a positive real part and the roots inside the unit circle distribute almost symmetrically around $\text{Re}(z)=0$.

To check we are in the second case, you may compute the number of zeroes in the strip $0<\text{Re}(z)<1$.

  • $\begingroup$ How can I check the zeros in that stripe? $\endgroup$ – Daniel Katzan Aug 30 '15 at 16:37
  • $\begingroup$ @DanielKatzan: you may compute the zeroes in $\text{Re}(z)\geq 1$ with the same Cayley-transform-technique, you just need an extra translation. $\endgroup$ – Jack D'Aurizio Aug 30 '15 at 16:47
  • $\begingroup$ It is also interesting to point out that you may prove: $$\int_{0}^{1}\frac{p'(z)}{p(z)}\,dz = \log\frac{17}{4}$$ only using Viète's theorem. $\endgroup$ – Jack D'Aurizio Aug 30 '15 at 16:52
  • $\begingroup$ I only know the Argument priniciple and Roches thereoem, we havent learned the Cayley-transform, and this is a hw question, so Im looking for a different method to solve this $\endgroup$ – Daniel Katzan Aug 30 '15 at 17:22

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