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$\newcommand{\cis}{\operatorname{cis}}$>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?

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    $\begingroup$ Hint: Substitute $w=z^4$. (If $z$ is a root on the unit circle, then $z^4$ will be as well---do you see why?) $\endgroup$ – Semiclassical Mar 31 '17 at 14:08
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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?

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  • $\begingroup$ 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 $1$? $\endgroup$ – gbox Mar 31 '17 at 14:14
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    $\begingroup$ If $|z|=1$ then $|f(z)|\ge101-1-3-1=96$. $\endgroup$ – Julián Aguirre Mar 31 '17 at 14:17
  • $\begingroup$ 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 $\endgroup$ – gbox Mar 31 '17 at 14:18
  • $\begingroup$ the product of the roots is 1 by Vieta's formulas? $\endgroup$ – gbox Mar 31 '17 at 14:24
  • $\begingroup$ Yes, but you can see it directly from $f(z)=(z-z_1)\dots(z-z_{12})$, where $z_i$ are the zeros of $f$. $\endgroup$ – Julián Aguirre Mar 31 '17 at 14:26
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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.

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  • $\begingroup$ I think you mean $g(z)=101z^4$. $\endgroup$ – Clayton Mar 31 '17 at 14:24
  • $\begingroup$ oh yes indeed thanks :) $\endgroup$ – Jennifer Mar 31 '17 at 14:24
  • $\begingroup$ Why did you take $g(z)=101z^4$ and not the higher powers? $\endgroup$ – gbox Mar 31 '17 at 14:36
  • $\begingroup$ Because it was a polynomial $g$ such that for all $|z|\leq1$, $|f(z)-g(z)|<|g(z)|$. $\endgroup$ – Jennifer Mar 31 '17 at 14:40

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