7

$\newcommand{\cis}{\operatorname{cis}}$>Prove that $$f(z)=z^{12}+3z^8+101z^4+1$$ has a root on the unit circle or $|z|\leq 1$

So started with looking at $$z^{12}+3z^8+101z^4+1=0$$

Therefore $$z^{12}+3z^8+101z^4=-1$$

looking at $z=r\cis\theta$ we get

$$r^{12}\cis(12\cdot\theta)+3r^8\cis(8\cdot \theta)+101r^4\cis(4\cdot \theta)=-1$$

And I can see that if $x=r^4\cis(4\theta)$ we get

$$x^3+3x^2+101x+1=0$$

I also know that if $z$ is a solution so is $\overline{z}$

How should I continue?

Moreover: Can we say that $12$ degree polynomial as $12$ complex roots ($z$ and $\overline{z}$) but because we have $z^8$ and $z^4$ so there will be less than $12$ solutions?

Michael Hardy
  • 1
gbox
  • 12,867

2 Answers2

4

It is easy to see that there are no roots on the boundary of the circle. Moreover, the product of the roots is $1$. How can this be?

Julián Aguirre
  • 76,354
  • Sorry for the qeustion but I am just a second year student. Why there are no roots on the boundary? I know that if $|z|=1 \iff z=\frac{1}{\overline{z}}$ is it have something to do with this fact?

    Why the product of the roots is $1$?

     – gbox Mar 31 '17 at 14:14
  • 2
    If $|z|=1$ then $|f(z)|\ge101-1-3-1=96$. – Julián Aguirre Mar 31 '17 at 14:17
  • Ok, it took me some time but I got it (thanks to semicalssical comment) if the root is on the boundary so does the root to the n-th power – gbox Mar 31 '17 at 14:18
  • the product of the roots is 1 by Vieta's formulas? – gbox Mar 31 '17 at 14:24
  • Yes, but you can see it directly from $f(z)=(z-z_1)\dots(z-z_{12})$, where $z_i$ are the zeros of $f$. – Julián Aguirre Mar 31 '17 at 14:26
  • So if $|x_{i}|\neq$ 1 and$ x_{1}\times x_{2}\times...\times x_{n}=1 \Rightarrow \exists x_{i}<1$ – gbox Mar 31 '17 at 14:27
1

You can apply Rouché's theorem. Let $f(z)=z^{12}+3z^8+101z^4+1$ and $g(z)=101z^4$.

For $|z| \leq 1$ : $$|f(z)-g(z)|=|z^{12}+3z^8+1|\leq|z^{12}|+|3z^8|+|1|=5$$ $$|g(z)|=101$$

We can apply the theorem because $5<101$. So $f$ has the same numbers of roots as $g$ in $\{z\in \mathbb{C};|z|\leq1\}$, so $f$ has $4$ roots in the domain considered.

Bérénice
  • 9,367