The number of solutions of $z^5+2z^3-z^2+z=a$ for $a\in \mathbb{R}$

Question

How we can calculate the number of solutions of $$z^5+2z^3-z^2+z=a\;\;,\;\;a\in \mathbb{R}$$ in the half-plane $\mathfrak {Re}(z)\ge 0$.

Any hint would be appreciated.

Answer 1

Since the locations of the roots in $\Bbb{C}$ will vary continuously with $a$, it suffices to figure out the values of $a$ at which they cross the imaginary axis and check some specific values in between. It will be helpful to notice that there is always exactly one real root, since $f'(x)=5x^4+5x^2+(x−1)^2>0$. The lone real root has the same sign as $a$.

Suppose the equation has roots of the form $ki$, for $k \in \Bbb{R}$. Then $k^5i-2k^3i+k^2+ki=a$; equating real and imaginary parts gives $$ k^2=a \\ k^5-2k^3+k=0 $$ The second equation factors as $k(k^2-1)^2=0$; so we must have $k=0$ or $k=\pm 1$, and the critical values of $a$ are $a=0$ and $a=1$.

Now, notice:

• When $a=0$, the four complex roots must sum to zero by the Vieta formulae. The above calculation shows that none of the complex roots can be pure imaginary; since they come in conjugate pairs, it follows that two must have positive real part and the other two negative real part. So there are three total roots with non-negative real part.
• Slightly moving $a$ away from $0$ will not affect which half-plane those two conjugate pairs of roots are in. Since it will affect the sign of the real root, the equation has two roots with non-negative real part for $a=-\epsilon$ sufficiently small (and hence for all negative $a$) and three such roots for $a=\epsilon$ (and hence for all $a \in (0,1)$).
• When $a=1$, the quintic factors as $(z^2+1)(z^3+z−1)$. The $z^3+z-1$ factor includes the real root of the original equation, which must be positive as $a$ is; since its roots again sum to zero, its two complex roots will have negative real part. So there are three roots with nonnegative real part (the positive real root and $\pm i$).
• Finally, for very large $a$, roots all need to have large magnitude. And for large input, the mapping $z\mapsto z^5+2z^3−z^2+z$ is dominated by the $z^5$ term. Under the simpler mapping $z\mapsto z^5$, positive reals have a preimage located properly in each of the four quadrants (plus one real preimage). So for very large $a$, we reason there will be a root in each of the two right quadrants, and a root in each of the two left quadrants. This gives us three roots with nonnegative real part in total (counting the real one).

Putting all this together, the equation has two solutions with nonnegative real part when $a$ is negative, and three otherwise.