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I've been doing some problems from Algebra and Geometry by Alan Beardon but I can't seem to do this one:

Suppose for $n\geq2$, show that the roots of $z^n+nz+1=0$ lies inside the circle $|z|=1+\frac{2}{n-1}$

The question is similar to this one. I can't seem to apply the reasoning in the answer to that question to this one, even though they are really quite similar. In my attempts I've tried using proof by contradiction and so assuming: $$ z^n+nz+1=0 \; \land \; |z|>1+\frac{2}{n-1}=\frac{n+1}{n-1}$$ and then tried to get to a contraction, though I can't seem to. In line with the similar question I tried applying the Bernoulli inequality (since $|z|\in\mathbb{R}$ and $n\geq2$): $$ |z^n|=(1+(|z|-1))^n>1+(|z|-1)n $$ or potentially using the triangle inequality: $$ z^n=-nz-1 \rightarrow|z^n|=|nz+1|\leq n|z|+1 $$ but no luck so far.


Any hints as to which direction to take would be great, I seem to be going round in circles a little bit!

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  • $\begingroup$ Try using Roche's theorem! $\endgroup$
    – Koro
    Sep 23, 2015 at 14:17
  • $\begingroup$ @kilimanjaro it seems from looking it up that that's the standard way to do this type of question. It's just that Roucher's theorem isn't introduced and so I get the impression the author wants you to do it without it. Thanks $\endgroup$
    – Jay
    Sep 23, 2015 at 14:28
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    $\begingroup$ It may be the case to notify the author that the problem statement is simply wrong. It is true if we replace $1+\frac{2}{n-1}$ with $n^{\frac{1}{n-1}}+\frac{1}{n^2-n}$, as shown below. $\endgroup$ Sep 23, 2015 at 15:25

2 Answers 2

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Assuming $z=\rho e^{i\theta}$, we have:

$$ \left\| z^n + n z\right\|^2 = \rho^{2n}+n^2 \rho^2 + 2n \rho^{n+1}\cos((n-1)\theta)\tag{1} $$ hence: $$ \left\| z^n + n z\right\|^2 \geq \rho^{2n}+n^2 \rho^2-2n\rho^{n+1}=(\rho^n-n\rho)^2.\tag{2}$$ We may notice that $g(\rho)=\rho^n-n\rho$ is a convex function by computing its second derivative. The only positive root of such a function is at $\rho=n^{\frac{1}{n-1}}$ and in such a point we have $g'(\rho)=n^2-n$. By convexity we have that, if

$$\|z\|> n^{\frac{1}{n-1}}+\frac{1}{n^2-n},\tag{3}$$ $z$ cannot be a root of the given polynomial.

Unluckily, if $n\geq 4$ this bound is slightly weaker than the original bound $1+\frac{2}{n-1}$.

It is not surprising, since the other answer clearly shows that the original bound does not hold.

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    $\begingroup$ Thanks for the answer. This seems much more complicated than the previous part. Probably owing to the fact that the question asks to prove something that's not true! $\endgroup$
    – Jay
    Sep 23, 2015 at 15:26
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I ran this in Mathematica:

Abs[z /. NSolve[z^5 + 5 z + 1 == 0, z]]

and I got this answer:

{1.46103, 1.46103, 0.199936, 1.53071, 1.53071}

so two of the roots lie outside the disk $|z|\leq 1.53$, therefore cannot lie inside the disk $|z|<1+{2\over 5-1}=1.5$

So it would seem what you asked to prove is not true.

Added:

Now that we are convinced that the original formulation is wrong, it might be the case that there is a silly typo here, and instead of $z^n+nz+1$ we should have $z^n+2z+1$. Indeed, taking $f(z)=z^n$ and $g(z)=z^n+2z+1$, we have $|f(z)-g(z)|<|f(z)|$ on the boundary $|z|=1+{2\over n-1}$, because the difference is $|2z+1|$, and it is bounded by $3+{4\over n-1}$ on the circle $|z|=1+{2\over n-1}$, whereas for such $z$: $$|z|^n=(1+{2\over n-1})^n>1+{2n\over n-1}+{n\choose 2}{4\over (n-1)^2}>3+{2n\over n-1}\geq 3+{4\over n-1}$$ and here we also get to use $n\geq 2$. So Rouche's theorem applies.

This makes some sense because from the reference pointed out by the OP we know that $z^n+z+1$ has no roots outside the disk $|z|\leq 1+{1\over n-1}$, so it seems a nice variation to put $2$ instead of $1$, both in the radius and in the coefficient of $z$. This of course leads one to wonder whether perhaps $z^n+kz+1$ has no roots outside the circle $|z|=1+{k\over n-1}$. There is good reason to believe so, because applying the Rouche's-theorem argument above for $k$ will bound $|f-g|$ on the circle's boundary by the amount $k+1+{k^2\over n-1}$, whereas the value of $z^n$ on that boundary, namely $(1+{k\over n-1})^n$, reminds us of $e^k$, which ought to be larger than a polynomial expression in $k$. Interesting, yet seemingly solvable problem. I might try to fill in the details later.

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  • $\begingroup$ Interesting. However, my weaker bound $1.54535$ works. $\endgroup$ Sep 23, 2015 at 15:14
  • $\begingroup$ Also fails for $n=6$, but $5$ was the first one. $\endgroup$ Sep 23, 2015 at 15:15
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    $\begingroup$ Thanks! That would explain why it's been difficult to prove. It's false! $\endgroup$
    – Jay
    Sep 23, 2015 at 15:24
  • $\begingroup$ Was that a problem given in a textbook?? $\endgroup$ Sep 23, 2015 at 15:25
  • $\begingroup$ @uniquesolution Yes, it was the second part to the other question that I linked to in the question. I'll write to the publishers to see what the solutions manual says about it, they have it locked to only give access to teachers. $\endgroup$
    – Jay
    Sep 23, 2015 at 15:31

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