I am working on an exercise stating as
How many roots of $z^{5}-4z^{3}+6=0$ lie in the closed quadrant $\Im(z)\geq 0$ and $\Re(z)\geq 0$ of the complex plane. How many roots are in the open quadrant?
It is relatively straight forward for the second question. We can apply the Rouche's theorem to $f(z):=z^{5}+6$ and $g(z):=-4z^{3}$, and consider the the contour $K$ as the $1/4$ circle with $R$ big enough such that all roots of $p(z):=z^{5}-4z^{3}+6$ lie in $|z|\leq R$. Then we have on the boundary $\partial K$, we always have $|f(z)|>|g(z)|$, so we know that $f(z)$ and $p(z)$ have the same number of roots inside $K$, then we just compute the solution of $$z^{5}=-6,$$ which only has one root in the first quadrant, since the distribution of the roots depends on the fifth roots of unity.
However, I have no idea about how to compute the roots in the closed quadrant, since neither Rouche nor Argument principle applies along the boundary.
What should I do? Thank you!
Edit 1:
Well I do not think the second question is that straightforward anymore since it is hard to argue $|f(z)|>|g(z)|$, if we have $z^{5}-z^{3}+6=0$, it is quick, but now we have a coefficient $4z^{3}$.
What should I do?
Best Answer
To get the roots in the closed quadrant you need to find out whether any real or pure imaginary roots exist. Such roots become real when they are squared.
Apply root squaring:
$z^5-4z^3=-6$
$(z^2)^5-8(z^2)^4+16(z^2)^3=36$
$(z^2)^5-8(z^2)^4+16(z^2)^3-36=0$
Find the critical points of the root-squared polynomial:
$5(z^2)^4-32(z^2)^3+48(z^2)^2=0$
$z^2\in\{0,2.4,4\}$
Numerical calculation reveals that
$(0)^5-8(0)^4+16(0)^3-36<0$
$(2.4)^5-8(2.4)^4+16(2.4)^3-36<0$
$(4)^5-8(4)^4+16(4)^3-36<0$
The critical points in $z^2$ show no sign changes in the root-squared polynomial so there are no real roots between them; the only real root for $z^2$ is the one lying outside the range of critical points due to the polynomial having odd degree. This root will satisfy $z^2>4$ so $z$ itself is real instead of pure imaginary.
We must now assess whether the real root is positive or negative. Descartes Rule of signs identifies an even number of positive roots and an odd number of negative ones, so the single real root in $z$ is negative meaning not in the closed first quadrant.
We conclude that there is no root on the boundary of the quadrant and thus one root in the closed quadrant overall.