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?

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
$$\boxed{\text{ on |z|=1}}.$$
