How many roots of $f(z) = z^3 +cz^2 + z + 1$ lie inside $|z|<1$ if
Part 1) $c>3$
Part 2) $3\ge c >2$
Part 3) $2\ge c >1$
Part 4) $c=1$
I am able to solve Parts (1) and (4) by Rouche's Theorem. Part (1) is trivial. Part (4) required a bit more work, and choosing $g(z)$ as $cz^2+z+1$, which reduced to just $z^2 + z+1$, when c =1. The roots of g(z), found by using the quadratic formula, lie on the circle and not inside, and hence $f(z)$ has no roots inside the disk either.
But I do not know how to solve Part (2) and Part (3).
Any ideas are welcome.
Thanks,
The zeros move continuously with $c$, so you have to find the cases when $z= e^{i \xi}$ with $\xi$ real is a solution. So we get the set of equations (real and imaginary part of your equation) $$\cos (\xi )+ c \cos (2 \xi )+\cos (3 \xi )+1=0, \qquad \sin (\xi )+c \sin (2 \xi )+\sin (3 \xi )=0.$$
Assuming $\xi \not \in \{-\pi/2,0, \pi/2, \pi \}$, we can eliminate $c$ from the second equation with the result $$ c= - \frac{\sin(\xi) + \sin(3 \xi)}{\sin(2 \xi)}.$$ Plugging this result in the first equation yields the contradiction $1=0$.
So the only possible places where a root can enter the unit circle is at $z=\pm 1,\pm i$.
In order to find the cases where a root enters we try the four cases
$\xi =0$, $z=1$ then $c=-3$.
$\xi =\pi$, $z=-1$ then $c=1$.
$\xi = \pi/2$, $z=i$ then $c=1$.
$\xi = -\pi/2$, $z=-i$ then $c=1$.
So the case $c=1$ is special and you have already treated this yourself. For $c>1$ there is never a zero on the unit circle, so the number of zeros inside the unit disk stays constant equal to 2 (as you have already shown in part 1).
To completeness here the discussion of all values of $c$:
The discussion of the four cases showed that for $c=1$ there are all three roots on the unit circle (so the number of zeros inside is 0). This is a very special point because taking the absolute value of Vieta's theorem $|z_1|\, |z_2 |\, |z_3| =1$ shows that otherwise (if not all three of them are on the circle) 2 or 1 are inside.
For $c>1$ the number of zeros cannot change, so we can discuss the case $c\to \infty$. In this case we have $f(z) \sim c z^2$ and thus two zeros inside the unit circle.
Next we treat the case $c<-3$. For $c\to -\infty$, we have $f(z) \sim c z^2$ and thus also two zeros inside the unit circle.
For $c=3$ there is 1 zero on the unit circle (at $z=1$), see cases above. We easily can solve the remaining two zeros and obtain that there is 1 zero inside the unit circle.
The only remaining case is $-3<c<1$. We can either solve 1 case, e.g., $c=0$. Or we can do perturbation theory for the zero which is on the unit circle in $\delta c= c+3$. We then see that this zero moves from inside the unit circle (for $c<-3$) to outside the unit circle (for $c>-3$).
In conclusion we have the number of zeros inside the unit circle given as $$ \begin{cases} 2, & c<-3 \lor c>1,\\ 1, & -3 \leq c < 1,\\ 0, & c=1. \end{cases}$$