Let \(n\) be a positive integer and let \(\alpha,\beta\in\mathbb R\setminus\{0\}\). Prove that the number of roots, counted with multiplicity, of \[ z^{2n}+\alpha^2z^{2n-1}+\beta^2=0 \] having positive real part is
\(n\) if \(n\) is even;
\(n-1\) if \(n\) is odd.
Solution
Set \[ P(z) = z^{2n} + \alpha^2z^{2n-1} + \beta^2. \]
1. The polynomial \(P\) has no zero on the imaginary axis.
For \(y\in\RR\), \[ \begin{aligned} P(iy) &= (-1)^n y^{2n} + \alpha^2(-1)^{n+1}i\,y^{2n-1} + \beta^2\\ &= \bigl( \beta^2+(-1)^ny^{2n} \bigr) + i(-1)^{n+1}\alpha^2y^{2n-1}. \end{aligned} \] If \(P(iy)=0\), its imaginary part vanishes. Since \(\alpha\neq0\), this forces \(y=0\). But then \[ P(0)=\beta^2\neq0. \] Thus no such zero exists.
2. For all sufficiently large \(R\), every zero of \(P\) lies in\[ \{z:\abs{z}<R\}, \]and the positively oriented boundary of the right half-disk consists ofthe semicircle\[ \Gamma_R: z=Re^{i\theta}, \qquad -\frac\pi2\leq\theta\leq\frac\pi2, \]followed by the imaginary segment\[ I_R: z=iy, \qquad R\geq y\geq-R. \]
The polynomial has finitely many zeros, so choose \(R\) larger than the modulus of every zero. Step 1 shows that no zero lies on the imaginary segment. The stated orientation keeps the right half-disk on the left while traversing its boundary.
3. Let \(N_+\) be the number of zeros of \(P\) with positive real part,counted with multiplicity. For every sufficiently large \(R\),\[ 2\pi N_+ = \Delta_{\Gamma_R}\arg P + \Delta_{I_R}\arg P. \]
For \(R\) as in step 2, the boundary contains no zero of \(P\), and the right half-disk contains exactly the zeros with positive real part. The argument principle says that the total change of a continuous argument of \(P\) around this positively oriented boundary equals \(2\pi\) times the number of enclosed zeros, counted with multiplicity.
4. Along the semicircle,\[ \Delta_{\Gamma_R}\arg P \longrightarrow 2n\pi \]as \(R\to\infty\).
For \(z\in\Gamma_R\), \[ P(z) = z^{2n} \left( 1+\frac{\alpha^2}{z} +\frac{\beta^2}{z^{2n}} \right). \] The factor in parentheses converges uniformly to \(1\) on \(\Gamma_R\) as \(R\to\infty\). For sufficiently large \(R\), it lies in the disk \[ \{w:\abs{w-1}<1/2\}, \] which admits a continuous argument converging uniformly to \(0\). Hence the argument change contributed by this factor tends to \(0\).
Meanwhile, on \[ z=Re^{i\theta}, \qquad -\frac\pi2\leq\theta\leq\frac\pi2, \] the factor \(z^{2n}\) has continuous argument \(2n\theta\), whose change is \[ 2n \left( \frac\pi2-\left(-\frac\pi2\right) \right) = 2n\pi. \] Adding the two argument changes proves the limit.
5. If \(n\) is even, then\[ \Delta_{I_R}\arg P \longrightarrow 0 \]as \(R\to\infty\).
When \(n\) is even, step 1 gives \[ P(iy) = \bigl(\beta^2+y^{2n}\bigr) -i\alpha^2y^{2n-1}. \] Its real part is strictly positive for every \(y\). Thus the path \[ y\longmapsto P(iy) \] lies entirely in the open right half-plane, where the argument has a single continuous branch taking values in \[ \left(-\frac\pi2,\frac\pi2\right). \]
As \(y\to\pm\infty\), \[ \frac{\operatorname{Im}P(iy)} {\operatorname{Re}P(iy)} = \frac{-\alpha^2y^{2n-1}} {\beta^2+y^{2n}} \longrightarrow 0. \] Hence the arguments at both endpoints of the downward segment \(y=R\) to \(y=-R\) tend to \(0\). Their difference therefore tends to \(0\).
6. If \(n\) is odd, then\[ \Delta_{I_R}\arg P \longrightarrow -2\pi \]as \(R\to\infty\).
When \(n\) is odd, \[ P(iy) = \bigl(\beta^2-y^{2n}\bigr) + i\alpha^2y^{2n-1}. \] For \(y>0\), the imaginary part is positive, so the path lies in the open upper half-plane. As \(y\to+\infty\), its argument tends to \(\pi\), because the real part is negative of order \(y^{2n}\) while the imaginary part has lower order \(y^{2n-1}\). As \(y\to0^+\), \[ P(iy)\longrightarrow\beta^2>0, \] so its argument tends to \(0\). Thus the argument change while traversing from \(+\infty\) down to \(0\) is \(-\pi\).
For \(y<0\), the imaginary part is negative, so the path lies in the open lower half-plane. As the traversal continues from \(0\) down to \(-\infty\), the argument changes from \(0\) to \(-\pi\). This contributes a second change of \(-\pi\). Hence the full downward imaginary-axis traversal has limiting argument change \[ -\pi-\pi=-2\pi. \]
7. If \(n\) is even, then\[ \boxed{ N_+=n. } \]
By steps <1>3–<1>5, \[ 2\pi N_+ = \Delta_{\Gamma_R}\arg P + \Delta_{I_R}\arg P \longrightarrow 2n\pi. \] The left-hand side is independent of \(R\) for all sufficiently large \(R\). Therefore \[ 2\pi N_+=2n\pi, \] so \(N_+=n\).
8. If \(n\) is odd, then\[ \boxed{ N_+=n-1. } \]
By steps 3, 4, and 6, \[ 2\pi N_+ \longrightarrow 2n\pi-2\pi = 2(n-1)\pi. \] Again the left-hand side is independent of sufficiently large \(R\), so \[ N_+=n-1. \]
9. Q.E.D.
Step 7 proves the even case, and step 8 proves the odd case.