How to solve this complex equation for the modulus of z? The question is as follows:

All the roots of the equation $11z^{10}+10iz^9+10iz-11=0$ lie:
  $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (i=\sqrt{-1})$
  (a) inside $|z|=1$
  (b) on $|z|=1$
  (c) outside $|z|=1$
  (d) can't say

The answer is (b). I tried factorizing it, but to no avail. Also, it doesn't appear to me that taking modulus would help. How to approach this problem?
 A: Keeping the higher order terms on the LHS and the lower order terms on the RHS, we have
$$z^9(11z+10i)=11-10iz$$
We first check that $11z+10i \neq 0$. Suppose on the contrary that it is, then $z=\frac{-10i}{11}$ and $LHS=0$, but $RHS=11-10i\left(\frac{-10i}{11}\right) \neq 0$. Hence, we can divide both sizes by $11z+10i$.
$$z^9=\frac{11-10iz}{11z+10i}$$.
Taking modulus and squaring both sides,
\begin{align}
\left|z\right|^{18}&=\frac{\left|11-10iz\right|^2}{\left|11z+10i\right|^2}\\
&=\frac{\left(11-10iz \right)\left(11+10i\bar{z} \right)}{\left(11z+10i \right)\left( 11\bar{z}-10i \right)}\\
&=\frac{121+110i\bar{z}-110iz+100\left| z\right|^2}{121 \left|z \right|^2+110i\bar{z}-100iz+100}\\
&= \frac{A}{B}
\end{align}
where I let the numerator expression in the second last line above be $A=121+110i\bar{z}-110iz+100\left| z\right|^2$  and the denominator in the second last line above be $B=121 \left|z \right|^2+110i\bar{z}-100iz+100.$
If $|z|>1$, then $|z|^{18}>1$, then $A>B$ and $A-B>0$, but $$A-B=121 \left( 1-\left| z\right|^2\right)+100\left( \left| z\right|^2-1\right)=21(1-\left| z\right|^2)$$
which is negative since $\left|z\right|^2>1$ by our assumption. Hence a contradiction.
Similarly, if $|z|<1$, $|z|^{18}<1$, then $A<B$ and $A-B<0$, but $A-B=21(1-|z|^2)>0$ which is again a contradiction.
Hence $|z|=1$.
A: Substitution $z=e^{it}$ gives the trigonometrical equation
$$11\sin5t+10\cos4t=0,\qquad(1)$$
or
$$\cos 4t=-1.1\sin 5t.$$
Easy to see that 
$$RHS\left(\dfrac{2k+1}{10}\pi\right)=1.1(-1)^{k+1}$$
for $k=-3,-2,-1,0,1,2$, so LHS and RHS have at least five intersections for $t\in\left(-\dfrac\pi2,\dfrac\pi2\right)$.  

This means that $(1)$ has at least 5 real roots for $t\in\left(-\dfrac\pi2,\dfrac\pi2\right)$.
On the other hand, it is known that
$$\sin5t=16\sin^5t-20\sin^3t+5\sin t$$
and
$$\cos4t=8\sin^4t-8\sin^2t+1,$$
so $(1)$ is equivalent to
$$176y^5+80y^4-220y^3-80y^2+55y+10=0,\quad y=\sin t.$$
In this way, the 5th order polynomial has $5$ real roots for $y\in(-1,1)$.
Therefore, equation $(1)$ has only real roots.
Thus, the right answer is
$$\boxed{\text{ on |z|=1}}.$$
