zeros of a polynomial Given $P(z)=z^6+6z+10$, find how many roots are in each quadrant  
I have already seen that $P(z)$ has six different roots, and that none of them are real or of the form $ki$, $k\in \Bbb R$.
Since the coefficients of $P(z)$ are real, then it's roots are 3 conjugate pairs, and therefore the possibilities are limited to two.
Defining $f(z)=z^6$, $g(z)=z^6+6z+10$ and applying Rouche in
$|z|<2$, $|f(z)-g(z)|<|f(z)|$
I get that all roots are contained in the disc, but how do I go about splitting them into the four quadrants? From what I know about this polynomial, finding zeros in one quadrant is enough to determine the rest, but I'm stuck trying to apply rouche to a quarter circle (basically I'm having trouble finding a suitable f and g)
 A: Rather than using Rouché's theorem we can approach this as a perturbation problem.  To wit, consider the polynomial
$$
P_a(z) = z^6 + az + 10.
$$
One can show that if $a \neq 0$ then $P_a$ has no zeros on the imaginary axis and if $0 \leq a < 6 \cdot 2^{5/6}$ then $P_a$ has no zeros on the real axis either.
If $a = 0$ then $P_a$ has six simple zeros -- one in each quadrant and one at each point $z = \pm 10^{1/6} i$.

By the inverse function theorem these zeros are analytic functions of $a$ for $a$ small enough.  If $z = z(a)$ is one of these zeros then
$$
z^6 + az + 10 = 0
$$
and, differentiating with respect to $a$,
$$
6z^5 z' + z + az' = 0,
$$
so that
$$
z' = - \frac{z}{a + 6z^5}.
$$
If we consider the zero with $z(0) = 10^{1/6} i$ then
$$
z'(0) = - \frac{1}{6 \cdot 10^{2/3}}.
$$
Consequently, the zero of $P_a$ located at $z = 10^{1/6} i$ when $a = 0$ moves into the left half-plane as $a$ increases past zero.  We remarked earlier that $P_a$ has no purely imaginary zeros if $a \neq 0$, so this zero must lie in the left half-plane for all $a > 0$.
Since the coefficients of the polynomial are real, the same is true for the zero located at $z = -10^{1/6} i$ when $a = 0$.

Each of the zeros starting in the four quadrants when $a = 0$ must remain in their quadrant for all $0 \leq a < 6 \cdot 2^{5/6}$, and once the zeros on the imaginary axis fall into the quadrants in the left half-plane they must remain there as well.
Taking $a = 6$ (and noting that $6 < 6\cdot 2^{5/6}$) we conclude that $P$ has two zeros in each of the quadrants II and III and one zero in each of the quadrants I and IV.  
