Using Rouche's Theorem to find the number of zeros of $p(z)=z^8 +10z^3 −50z+1$ in the right halfplane

Question

I'm studying for a complex analysis qualifying exam and was wondering if someone could help me out with this. I am not sure how to apply Rouche's Theorem to this.

How many zeros does the polynomial $$p(z)=z^8 +10z^3 −50z+1$$ have in the right half-plane?

Answer 1

I am also studying for a complex analysis qualifier.
Here is my attempt at this problem.

Given $f(z) = z^{8}+10z^{3}-50z+1$, consider $g(z) = z^{8}+1$.
For $g(z) = 0$, there are eight roots all located on the unit circle.
Four of these roots are on the right half-plane.

Now consider a closed curve $\gamma$ large enough to include \begin{equation*} \{\left| z \right| = 1, \text{Re(z)} \ge 0\}. \end{equation*}

For example, let $\gamma$ = $C_{R} \bigcup L_{R}$ be the half-disk with \begin{equation*} L_{R} = \left\{ z \Big\vert \text{Re}(z) = 0, -iR \le \text{Im}(z) \le iR \right\} \end{equation*} \begin{equation*} C_{R} = \left\{ z \Big\vert z = Re^{i\theta}, -\pi \le \theta \le \pi\right\} \end{equation*}

For $R \ge 2$ on $C_{R}$: \begin{alignat*}{2} \left| g(z) \right| &\ge \left| R^{8}e^{i8\theta} + 1\right| \\ &\ge R^{8} - 1 \\ &\ge 10R^{3} + 50R \\ &\ge \left| 10R^{3}e^{i3\theta} - 50Re^{i\theta} \right| \\ &= \left| f(z) - g(z) \right|. \end{alignat*}

Similarly, for $R \ge 2$ on $L_{R}$ with $z = x + iy$: \begin{alignat*}{2} \left| g(z) \right| &\ge \left| (iy)^{8} + 1\right| \\ &\ge y^{8} - 1 \\ &\ge 10y^{3} + 50y \\ &\ge \left| 10(iy)^{3} - 50(iy) \right| \\ &= \left| f(z) - g(z) \right|. \end{alignat*}

Then on $\gamma$, $\left| g(z) \right| \ge \left| f(z) - g(z) \right|$.
By Rouche's Theorem, we have that $f(z)$ has four roots inside $\gamma$ which implies $f(z)$ has four roots in the right half-plane.

Answer 2

The above solution is wrong, I think this question cannot be solved by Rouche theorem, the argument principle may help.

Answer 3

Use the argument principle. See p. 227-228 of Complex Analysis by Gamelin for an example. Determine the change in argument of $p(z)$ as you traverse along a path that encloses the right half-plane.

It is easy to see that $p(z)$ has no zeros on the imaginary axis. For $z=iy$, $$ p(iy) = y^8+1 + i(-10y^3-50y) = y^8 + 1 -10iy(y^2+5), $$ and the real part of this never vanishes. Consider the half-disk in the right half-plane of radius $R$. For $R$ large enough, this will contain all of the zeros in the right-half plane. If we examine the change in argument of $p(z)$ as we traverse along the boundary of this half-disk, it will tell us how many zeros of $p(z)$ it contains.

First consider the change in argument of $p(z)$ along the semicircular path $\Gamma_R$ of radius $R$ that goes from $-iR$ to $iR$ in the right half-plane. Along this path, the $z^8$ term dominates in $p(z)$ and thus $\text{arg}\,p(z)\approx 8\,\text{arg}\,z$. Since the change in argument of $z$ along this path is $\pi$, the change in argument of $p(z)$ along $\Gamma_R$ is approximately $8\pi$.

Now check the change in argument of $p(z)$ along the path on the imaginary axis from $iR$ to $-iR$. For large $R$, the approximate values of $p(-iR)$ and $p(iR)$ are $$ p(iR) \approx R^8 \approx p(-iR) \hspace{10mm}, $$ Since the real part of $p(iy)$ is never zero, this path does not cross the imaginary axis and the change in argument will be roughly zero along this segment.

The change in argument must be $2\pi$ times the number of zeros contained in the right half plane. Since the change in argument around the entire path is $8\pi$, there must be four zeros in the right-half plane.