Suppose z is any root of $11z^8 + 20 iz^7 + 10iz –22 = 0$. Then $S = |z|^2+| z|+ 1$ satisfies? 
Suppose z is any root of $11z^8 + 20 iz^7 + 10iz –22 = 0$. Then $S = |z|^2+| z|+ 1$ satisfies ?(A) $S \leq 3$ (B) $3 < S < 7$ (C) $7 \leq S < 13$ (D) $S \geq 13$

Where do I start? I cannot simplify $z^7=\frac {22-10iz}{11z+20i}$. This could be written as $\frac {z^7+1}{z^7-1}= \frac {(2-z)(11+10i)}{22-11z-10iz-20i}$ but I don't see how that is useful.
 A: Let $f=11z^8+20iz^7+10iz-22$.

Let $a,b$ be roots of $f$ with the least and greatest absolute values, respectively.

By Vieta's formula, the product of the roots of $f$ is $-2$, hence we must have $|b| > 1$ and $|a| < 2$.

From $|b| > 1$, we get 
$$3 =1^2+1+1 < |b|^2+|b|+1$$
which eliminates choice $(A)$.

From $|a| < 2$, we get 
$$|a|^2+|a|+1 < 2^2+2+1=7$$
which eliminates choices $(C)$ and $(D)$.

Hence, given the multiple choice context, the answer must be choice $(B)$.
A: This is your best attempt at an answer.
By Descarte's rule of signs, we have one positive and one negative real root. As you noticed, the function can be rewritten as $11(iz)^8 - 20(iz)^7 + 10(iz) - 22 = 0$. Substituting $z=-1$ gives $1$, so we have a root near $-1$. Substituting $-1$ gives $S=3$.
However, we can approximate the function as $11(iz)^8 - 20(iz)^7 = 0$, so $iz=\frac{20}{11}$. Dividing by $z$ will not change the magnitude, and so substituting gives $S \approx 6.12$.
The only option that can accommodate both of these options is option B.
