Roots of $z^{2n} + \alpha z^{2n -1} + \beta ^2$ I've been looking at a problem available here. The problem is:
Let $n$ be a natural number, and $\alpha$, $\beta$ nonzero reals. Show that the number of roots of $p(z) = z^{2n} + \alpha z^{2n -1} + \beta ^2$ in the right half plane is
\begin{cases}
n&&\text{ if } n \text{ is even} \\
n \pm 1 &&\text{ if } n \text{ is odd} 
\end{cases}
I have a solution which uses the Argument Principle. The contour consists of two pieces: the line segment on the imaginary axis from $Ri$ to $-Ri$, and the semicircle in the right half-plane which connects $-Ri$ to $Ri$. Evaluating each piece separately and passing to the limit as $R \to \infty$ gives the answer (the $\pm 1$ for the odd case appears because of the branch cut for log$()$ on the negative real axis). 
My question is this: Is there another way to do this which does not involve using the Argument Principle? I thought maybe you could use a conformal map from the open unit disk to the right half-plane in order to rewrite the problem in terms of zeros on the unit disk (and then use Rouche's Theorem), but I didn't succeed in getting it to work.
Thanks again!
 A: For even $n$, we can directly use Rouché's theorem with the function $f(z) = z^{2n}+\beta^2$, using the half-disk with centre $0$ and radius $R$ in the right half plane. If $R$ is large enough, it contains all zeros of $p$ and of $f$ in the right half plane, and on the semicircle we have $\lvert f(z)-p(z)\rvert < \lvert f(z)\rvert$, while on the segment of the imaginary axis, $f(z)$ is real and strictly positive, but $f(z) - p(z) = \alpha z^{2n-1}$ is purely imaginary.
Thus $f(z) + \lambda (p(z)-f(z))$ has no zeros on the boundary of the half-disk for any $\lambda \in [0,1]$, and Rouché's theorem asserts that $p$ has equally many roots in the half disk as $f$. For $R > \lvert\beta\rvert^{1/n}$, exactly half of the roots of $f$ lie in the right half-plane, so we have $n$ roots for $p$ in the right half plane.
For odd $n$, we can't use quite the same argument, since then $z^{2n}+\beta^2$ has two simple roots on the imaginary axis (and $n-1$ roots in either half-plane).
But $h_\lambda(z) = f(z) + \lambda(p(z)-f(z))$ still has no zeros on the boundary of the half-disk for $\lambda \in (0,1]$, since $\alpha z^{2n-1}$ is purely imaginary on the imaginary axis, so the number of zeros of $h_\lambda$ in the right half plane is independent of $\lambda\in (0,1]$.
For $\lambda \to 0$, the zeros of $h_\lambda$ converge to the zeros of $f$, and the question is whether the two purely imaginary zeros $\zeta^{\pm} = \pm i\lvert\beta\rvert^{1/n}$ of $f$ are reached from the left or the right half plane.
The coefficients of $h_\lambda$ are real, hence the non-real roots occur in conjugate pairs, and therefore the two roots $\zeta^\pm$ are both reached from the same half-plane. If they are reached from the left half-plane, $p$ has $n-1$ roots in the right half-plane, and if they are reached from the right half-plane, $p$ has $n+1$ roots in the right half-plane.
For very small $\lambda > 0$, one step of the Newton algorithm leads to the approximation
$$\zeta^\pm_\lambda \approx \zeta^\pm - \frac{\lambda\alpha}{2n}\cdot \frac{1}{1 + \frac{\lambda\alpha}{\zeta^\pm}\left(1-\frac{1}{2n}\right)}$$
of the zeros, which lies in the left half plane if $\alpha > 0$, and in the right if $\alpha < 0$.
