If $11z^8+20iz^7+10iz-22=0$,then show that $1<|z|<2$

Question

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.

Answer 1

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 $1 < |z| < 2$.

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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.

Answer 2

After this step

$$z^7=\frac {22-10iz}{11z+20i}$$

$\implies z^7=\dfrac{-i(10z+22i)}{11z+20i}$

Take Modulus on both sides

$\implies|z^7|=\bigg|\dfrac{10z+22i}{11z+20i}\bigg|$

Case $1$: Let $|z|<1\implies \bigg|\dfrac{10z+22i}{11z+20i}\bigg|<1$

Next step is square on both side and use $z\bar z=|z|^2$

$\implies |10z+22i|^2<|11z+20i|^2\implies(10z+22i)(10\bar z-22i)<(11z+20i)(11\bar z-22i)$

So after multiplication

$21 |z|^2>84\implies |z|>2$, Hence Contradiction. So $|z|$ cannot be greater than $1$.

Case $(2)$:

If $|z|>1$ and Follow the same step as we did in case $(1)$

We get $|z|<2$.

Hence $1<|z|<2$.