Let $n\in \mathbb{Z}_{\geq1}$ and $\alpha$ and $\beta$ be nonzero real numbers. Let $r$ be the number of roots of $f(z) = z^{2n}+\alpha^2z^{2n-1}+\beta^2$ that have positive real part. Prove that $r=n$ if $n$ is even and $r=n-1$ if $n$ is odd.
I used a semicircle contour with vertical diameter and was able to show that the argument contribution of the semicircle part of the contour is $2n\pi$. I got that when $n$ is even, the vertical segment is $f(iy) = y^{2n} + \beta^2 -i\alpha^2y^{2n-1}$. This does not contribute to argument change because the real part is always positive and the imaginary part is always negative (so it's only in QIV). Hence when $n$ is even, the number of zeros is $n$.
However, I didn't understand the solutions for when $n$ is odd. The function becomes $f(iy)=-y^{2n} + \beta^2+i\alpha^2y^{2n-1}$. Supposedly this has an argument change of $-2\pi$, but I don't know why and how to show it. In the past, what I have been told to do when finding argument change is to see whether the real and imaginary parts are positive or negative between critical points and see what quadrant the curve lands in. I just didn't know how to do that in this case!
Thanks in advance!