Show that $z^5 - z +16$ has two roots in the right half plane Show that the polynomial 
$$z^5 - z +16$$
has all of its roots in the region
$$\{z\in \mathbb{C} \; | \; 1< |z| < 2\},$$
and show that two of its roots have positive real part.
I have used Rouché's theorem to prove that all of its roots are in the above region. But I don't have any clue on how to show the second part, that two of the roots are in the right half plane.
 A: Use the argument principle on a large D-shaped loop traversed anti-clockwise.
let $P(z) = z^5-z+16$ and first watch $P(z)$ on the imaginary axis.
$P(iy) = 16 + i(y^5-y)$ so it stays in the right half-plane, and looking at limits when $y \to \pm \infty$ we see that its argument decreases by $\pi$ as $y$ goes down from $+\infty$ to $-\infty$.
Meanwhile, if you look at $P$ on a large half-circle in the right half plane, $P(z) \sim z^5$, so that when $z$ follows along the circle, its argument increases by $\pi$ and so $z^5$ and $P(z)$'s argument increase by $5\pi$.
In total the argument changes by $5\pi-\pi = 4\pi$ during the entire loop, so it must have two roots inside it.
A: Hint: Let $\mathcal{R}$ be the right half plane.


*

*Let $P(X)$ be a polynomial with real coefficients. Show that if $\alpha$ is a root of $P(X)$, then so is its complex conjugate $\bar{\alpha}$. 

*Show that $P(X)=X^5-X+16$ has only one real root which is negative. (You can use your already existing result to show all real roots are negative. Furthermore, by showing thta $P(X)$ is monotonic in $1<|x|<2$, you'll be able to seal the deal).

*Conclude that the roots of $P(X)$ are of the form $r\in \mathbb{R}$ negative, $\alpha, \bar{\alpha}$ and $\beta, \bar{\beta}$ all complex roots. Argue that $P(X)$ has either zero, two or four roots in $\mathcal{R}$. Also show that
$$P(X)=(X-r)(X^2-2AX+|\alpha|^2)(X^2-2BX+|\beta|^2)$$
where $A=\mathrm{Re}[\alpha]$ and $B=\mathrm{Re}[\beta]$.

*By expanding the above relation (keep an eye on $X^4$), Prove that
$$
r=-2(A+B)
$$
Then show that $A+B>0$, and therefore having zero roots in $\mathcal{R}$ is impossible.

*Again by expanding the above relation (keep an eye on $X^2$), and the previous part, show that
$$
|\alpha|^2 A + 
|\beta|^2 B + 4AB(A+B)=0
$$
Argue that it is impossible to have both $A>0$ and $B>0$. This finishes the job.
