TITLE: Roots of $z^{11}-2z^6+z^4+10i$ in upper half plane.
QUESTION [6 upvotes]: I have this problem:

How many times the polynomial $P(z)=z^{11}-2z^6+z^4+10i$ goes to zero in the upper half plane?

Obviously I need the zeros of $P(z)$, and I thought it would be useful the Rouché theorem. 
For that I thought it was a good idea to get a disc to bound the problem, and it was considering $\vert z\vert =2$ and $g(z)=z^{11}$. Now, with it I obtained that the zeros were inside that disc. My problem is how to count the number of zeros in $B_2(0)\cap R$ with $R=\{z\colon \Im(z)\geq 0\}$. Can anyone help me how to do that?

REPLY [6 votes]: Well, since there are still no answers let me try.
One the real line there are no roots and the image of the real line by $P(z)$ lies inside the line $\Im(z)=10$.
The standard work, consider a large semi-circle with diameter lying on the real line (center at the origin for convenience) an containing all the roots of $P$ in the upper half plane. Call the radius $R$. The image of the diameter we know where it is, an the image of the circular part is close to the image of $R^{11}e^{11\theta i}$ where $\theta$ travels $[0,\pi]$. So, for $\theta=0$ the image lies near the point $R^{11}+10i$, for $\theta=\pi$ the image lies near the point $R^{11}e^{11\pi i}+10i=R^{11}e^{\pi i}+10i$. There are $\theta=2pi/11, 4\pi/11, 6\pi/11, 8\pi/11$, and $10\pi/11$, a total of $5$ winds around the origin.
So, $5$ roots in the upper half-plane.
No thinking, only computation. No Rouche'.