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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.

José Carlos Santos
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Tapi
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  • Are you sure that you have made no typo? – Dr. Sonnhard Graubner Aug 04 '19 at 10:52
  • You can write $$11z^8-22+i(20z^7+10z)=0$$ – Dr. Sonnhard Graubner Aug 04 '19 at 10:55
  • @Dr.SonnhardGraubner Yes, I am. It can also be written as: $11(iz)^8-20(iz)^7+10iz -22=0$ since it is not mentioned that $z$ is real. – Tapi Aug 04 '19 at 10:56
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    This is a classic example of Rouche's Theorem. Consider circles around the origin of radii $3,7,13$, and consider $f=11z^8+20iz^7+10iz-22$, $g=-11z^8-20iz^7$. If on the radius of the circle, $|g|<|f|$, then $f$ and $f+g$ have the same number of roots inside the circle. – Rushabh Mehta Aug 04 '19 at 11:11
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    Substituting $z=i$ gives $-1$, so there is a root near $z=i$ and $S=3$. Therefore the answer must be either A or B. – Toby Mak Aug 04 '19 at 11:11
  • @DonThousand I'm not familiar with Rouche's Theorem, could you please elaborate? Also, this is a high school math problem and I don't think it is supposed to be solved using that. – Tapi Aug 04 '19 at 11:15
  • @Tapi Rouche's Theorem – Rushabh Mehta Aug 04 '19 at 11:22
  • From what competition is this problem from? – Toby Mak Aug 04 '19 at 11:24
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    Rouche's Theorem is a formal way of saying you only have to look at dominating terms to find roots of polynomials. – Rushabh Mehta Aug 04 '19 at 11:24
  • @TobyMak It's from KVPY 2018, organized by the Indian Institute of Science. Also, the answer is B. – Tapi Aug 04 '19 at 11:27
  • @DonThousand : While Rouché in principle can be used this way to find the number of roots in $1<|z|<2$, 7 of the 8 roots have an absolute value between $1.0045$ and $1.030$, making finding a suitable polynomial $g$ manually very difficult. – Lutz Lehmann Aug 04 '19 at 13:40
  • Near duplicate question about the same polynomial: https://math.stackexchange.com/q/2990850/115115 with answers successfully using Rouché cleverly using the coefficient structure to circumvent above mentioned problem. – Lutz Lehmann Aug 04 '19 at 13:47

3 Answers3

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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 $(B)$.

quasi
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2

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.

Toby Mak
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  • I have the solution paper, and you can notice that 1.5 minutes to answer each question. The solution needed is likely not very rigorous, and likely requires a lot of intuition and deduction to get to the correct option. – Toby Mak Aug 04 '19 at 11:43
  • Yes, a rigorous solution is not needed. I didn't get how you approximated the function. Where did the $10iz-22$ go? – Tapi Aug 04 '19 at 11:49
  • You keep all the higher order terms $(iz)^8$ and $(iz)^7$. The root will be close to the original, since the other terms are nothing compared to the higher order terms. – Toby Mak Aug 04 '19 at 11:51
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    Okay, and before substituting, you're dividing by $i$, right? – Tapi Aug 04 '19 at 11:55
  • The other dominant binom is $20iz^7-22$ which has roots close to the unit circle so that $S$ is close to $3$ for these roots. – Lutz Lehmann Aug 04 '19 at 13:31
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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 $1<|z|<2\implies \bigg(|z|+\dfrac{1}{2}\bigg)^2+\dfrac{3}{4}\in(3,7)$

mathophile
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