How many roots has the polynomial $z^{2015} + 8z^{12} + 1$ in $\{z \in \mathbb{C}: \Im(z) > 0$ and $\Re(z) > 0\}$? I have to calculate how many zeros of $p(z) = z^{2015} + 8z^{12} + 1$ lies in the first quadrant of the complex plane. I'm used to solve this kind of problems in some disk by estimation using Rouche Theorem, but now no disk is given.
I was trying to use $h(z) = -8z^{12} - 1$ and conclude that it has 3 zeros in the first quadrant, but I can not prove that $|p(z) + h(z)| < |p(z)| + |h(z)|$. Any ideas?
 A: $P(z)=z^{2015}+8 z^{12}+1=0$ has 2015 roots
Only one is real. Indeed first derivative 
$P'(z)=2015 z^{2014}+96 z^{11}=z^{11} \left(2015 z^{2003}+96\right)$
$P'(z)=0 $ at $z_1=\sqrt[2003]{-\dfrac{96}{2015}}\approx -0.998481$ and $z_2=0$
$P'(z)>0$ in the interval $z<-0.998481\lor z>0$ so we have $z_1$ as a local maximum and $z_2$ as a local minimum, therefore as $P(-1.003)\approx -408.9$ and $P(-1)=8$ for the intermediate values theorem there is one zero $z_0\approx -1.0011$
As $P(0)=1>0$ is the only minimum we can state that $z_0$ is the only real root of $P(z)$
Therefore there are $2014 $ complex roots. As the polynomial has real coefficients, for any root we have its conjugate so only half of them $1007$ have positive imaginary part. 
Now I am going to state something I'm not sure to prove properly and I hope that someone will do it better.
From Vieta's relations between roots and coefficient we have that the sum of the roots is zero and the product is $1$. This is true also considering the symmetric of $z$ wrt $\Im(z)$ axis for which the polynomial is, if $z=\rho e^{i\theta}$
$P(\rho e^{i(\theta+ \pi)})=-\rho^{2015} e^{2015 i \theta}+8 \rho^{12}e^{12 i t}+1=-z^{2015}+8z^{12}+1$
This is zero when 
$z^{2015}-8z^{12}-1=0$
and have the same sum zero and product $-1$
Therefore I conjectured that  roots with $\Re (z )>0$ are (almost) half the total, that is $504$
The roots of $P(z)=0$ having $\Re (z)>0 \land \Im (z)>0$ are $504$
Hope someone will prove better the last part of this result

A: Let $C_R$ be the large semicircle centered at origin, lying in the right half-plane with radius $R$. We only need to find how many times $f(z)$ winds around the origin as $z$ travels along $C_R$ for very big $R$.
Note that $f(ix) = -ix^{2015} + 8x^{12} + 1$. As we travel from $Ri$ to $-Ri$, the real part of $f(z)$ is always positive, but imaginary part changes from $-\infty$ to $+\infty$, the winding number up to now is $1/2$ counterclockwise.
Then we travel from $-Ri$ along the arc to $Ri$, when $R$ is large, $f(Re^{i\theta}) \approx R^{2015} e^{2015i\theta}$. When $\theta$ ranges from $-\pi/2$ to $\pi/2$, $e^{2015i\theta}$ winds around origin $2015/2$ times counterclockwise.
Adding the two numbers, we conclude that $f(z)$ has $1008$ zeroes with $\Re(z)>0$. Since $f(z)$ has no positive real root, there are $504$ roots with $\Im(z)>0,\Re(z)>0$.
A: Set $h(z)=\frac18z^{2003}$ so that
$$
f(z)=p(z)+h(z)=(z^{2003}+8)(z^{12}+\tfrac18)
$$
Now one needs to check that $|h(z)|<|f(z)|$ along the imaginary coordinate axis $z=iy$, that is,
$$
\tfrac18|y|^{2003}<\sqrt{y^{4006}+64}·(\tfrac18+y^{12}).
$$
This is obviously true. 
Thus $p(z)$ and $f(z)$ have the same number of roots in the half-plane $Re(z)>0$. There are $6$ roots from the factor $(\tfrac18+y^{12})$ and $1002$ from the factor $(z^{2003}+8)$, for symmetry reasons there has to be an odd number in the negative half-plane. By again excluding the roots on the positive half-axis, one finds $504$ roots in the first quadrant.
