<s>[INST] If $11z^8+20iz^7+10iz-22=0$,then show that $1<|z|<2$ If $$11z^8+20iz^7+10iz-22=0$$then show that $$1<|z|<2$$
My Attempt:
If $z=x+iy$; then$$z^7=\frac{22z-10iz}{11z+20i}$$
$$|z|^7=\sqrt{\frac{400+440y+100(x^2+y^2)+84}{400+440y+100(x^2+y^2)+21(x^2+y^2)}}$$
If $x^2+y^2=4$ i.e. $|z|=2$
then $|z|^7=1$
which is clearly a contradiction.
Not able to proceed from here.
 [/INST] Let $p(z) = 11z^8+20iz^7 + 10iz - 22$.
The part for $|z|<2$ is already in the other answer, but I include it for the sake of completeness.  When $|z|=2$,
$$|20iz^7 + 10iz - 22| < 20(2^7)+10(2)+22 \\
< 21(2^7) < 22(2^7) = 11(2^8) = |11z^8|.$$
$11z^8$ has eight zeros (counted as many times as its multiplicity) in the open disk $|z| < 2$, so by Rouché's Theorem, $p$ has eight roots in $|z| < 2$.  Since $\deg(p) = 8$, all roots of $p$ lies inside $|z| < 2$.
To show that all roots of $p$ lies outside $|z| < 1$, apply Rouché's Theorem on $f(z) := 20iz^7 - 22$ and $g(z) := 11z^8 + 10iz$.  Observe that $p = f+g$, $f(z) = 2(10iz^7-11)$ and $g(z) = z(11z^7+10i)$, and that all roots of $f$ lies outside $|z| \le 1$.
$$f(z) = 0 \iff 20iz^7=22 \implies |z^7| = 1.1,$$
so on $|z| = 1$, $f(z) \ne 0$ and
$$|g(z)| = |11z^7+10i| = |11z^{-7}-10i| \\ = |11-10iz^7| = \frac{|f(z)|}{2} < |f(z)|,\tag{*}\label{*}$$
so all roots $z$ of $p$ satisfy $1 \le |z| < 2$, and it remains to show that $|z| \ne 1$.
Suppose $|z|=1$ and $p(z) = 0$.  $p = f + g$ by construction, so
$|f(z)| = |g(z)|$.  But $|f(z)| = 2|g(z)|$ from \eqref{*}, so $|f(z)| = |g(z)| = 0$.  When $f(z) = 0$, $z^7 = -11i/10$, so $1 = |z^7| = 11/10$, which is absurd.  Therfore, the unit circle $|z| = 1$ does not contain any root of $p$.
Hence, all roots of $p$ lie in the annular region $1 < |z| < 2$.

OP's reformulation $$z^7=\frac{22z-10iz}{11z+20i} \tag1\label1$$ can be hardly applied to solve this problem.  Let $z = x+yi$ with $x,y \in \Bbb{R}$.  
$$\begin{aligned}
|z|^7 &= \left\lvert\frac{22z-10iz}{11z+20i}\right\rvert \\
1 &= \left\lvert\frac{(22x+10y)+(22y-10x)i}{11x + (11y+20)i} \right\rvert \\
&= \sqrt{\frac{584(x^2+y^2)}{121(x^2+y^2)+440y+400}} \\
&= \sqrt{\frac{584}{521+440y}} \\
y &= \frac{63}{440}, x = \pm\sqrt{1-y^2}
\end{aligned}$$
Check that $x+yi$ does not verify \eqref{1} to conclude.
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