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