Suppose z is any root of $11z^8 + 20 iz^7 + 10iz –22 = 0$. Then $S = |z|^2+| z|+ 1$ satisfies?

Question

Suppose z is any root of $11z^8 + 20 iz^7 + 10iz –22 = 0$. Then $S = |z|^2+| z|+ 1$ satisfies ?
(A) $S \leq 3$ (B) $3 < S < 7$ (C) $7 \leq S < 13$ (D) $S \geq 13$

Where do I start? I cannot simplify $z^7=\frac {22-10iz}{11z+20i}$. This could be written as $\frac {z^7+1}{z^7-1}= \frac {(2-z)(11+10i)}{22-11z-10iz-20i}$ but I don't see how that is useful.

Answer 1

This is your best attempt at an answer.

By Descarte's rule of signs, we have one positive and one negative real root. As you noticed, the function can be rewritten as $11(iz)^8 - 20(iz)^7 + 10(iz) - 22 = 0$. Substituting $z=-1$ gives $1$, so we have a root near $-1$. Substituting $-1$ gives $S=3$.

However, we can approximate the function as $11(iz)^8 - 20(iz)^7 = 0$, so $iz=\frac{20}{11}$. Dividing by $z$ will not change the magnitude, and so substituting gives $S \approx 6.12$.

The only option that can accommodate both of these options is option B.

Answer 2

Let $f=11z^8+20iz^7+10iz-22$.

Let $a,b$ be roots of $f$ with the least and greatest absolute values, respectively.

By Vieta's formula, the product of the roots of $f$ is $-2$, hence we must have $|b| > 1$ and $|a| < 2$.

From $|b| > 1$, we get $$3 =1^2+1+1 < |b|^2+|b|+1$$ which eliminates choice $(A)$.

From $|a| < 2$, we get $$|a|^2+|a|+1 < 2^2+2+1=7$$ which eliminates choices $(C)$ and $(D)$.

Hence, given the multiple choice context, the answer must be choice $(B)$.

Answer 3

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 $1<|z|<2$.

Now According to Question

$|z|^2+|z|+1=\bigg(|z|+\dfrac{1}2{}\bigg)^2+\dfrac{3}{4}$

We have $1<|z|<2\implies \bigg(|z|+\dfrac{1}{2}\bigg)^2+\dfrac{3}{4}\in(3,7)$