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I am trying to determine how many roots $$p(z)=2z^4-3z^3+3z^2-z+1=0$$ has in the first quadrant.

My attempt:

We first note that $p(z)$ has real coefficients. Thus by the conjugate root theorem, the roots of $p(z)$ occur in complex conjugate pairs.

Next, we determine if $p(z)$ has roots on the axes.

Case 1: Suppose $p(z)$ has a root on the real axes, then $p(x)=2x^4-3x^3+3x^2-x+1=0$ for some $x\in\mathbb{R}$. Now by the rational root theorem, the only possible roots that $p(x)$ can have is when $x=1$. But $p(1)\neq 0$ and hence $p(z)$ cannot have a root on the real axes by contradiction.

Case 2: Suppose $p(z)$ has a root on the imaginary axes, then $p(iy)=2y^4+3iy^3-3y^2-iy+1=0$ for some $y\in\mathbb{R}$. Then, $$\Im(p(iy))=0\implies y(3y^2-1)=0.$$ But, $p(0)\neq 0$ and $p\left(\pm\frac{1}{\sqrt{3}}\right)\neq 0$. Hence $p(z)$ does not have a root on the imaginary axes, by contradiction.

So, the roots are $a\pm ib$ and $c\pm id$ for $a,b,c,d\in\mathbb{R}$ and $a,b,c,d\neq 0$. Now the sum of the roots equals zero, which implies $a=-c$. Hence, there is only one root in each quadrant.

My questions are, is my explanation of case $1$ and case $2$ correct? Also, (a very basic question) why does the sum of the roots equal zero?

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Your explanation for the case 1 is wrong. The rational root theorem is, as its name suggests, about rational roots. It doesn't allow you to deduce that there are no real roots (byt the way: in fact, there aren't). The explanation for case 2 is correct.

Also, the sum of the roots is not $0$; it is $-\dfrac{-3}2=\dfrac32$.

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  • $\begingroup$ For case $1$, if I divide through by $2$ and hence made the polynomial monic, could I use the integer root theorem? I'm unsure of how it is best to show that no real roots exist (I vaguely remember a theorem concerning the factors of the constant term of a polynomial). I alternatively thought of trying to show that $p(x)\geq 0$, but this seemed too difficult. Also, I do agree that the sum of the roots is $-\frac{b}{a}=\frac{3}{2}$. But what does this tell us about the placement of the roots in the complex plane? $\endgroup$ – JulianAngussmith Oct 23 '18 at 8:20
  • $\begingroup$ @JulianAngussmith, how do you imagine proving no integer roots (which follows from not having rational roots, by the way) will help you prove that there are no real roots? $\endgroup$ – Ennar Oct 23 '18 at 8:23
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    $\begingroup$ @JulianAngussmith Did I or did I not answer your original question? $\endgroup$ – José Carlos Santos Oct 23 '18 at 8:51
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    $\begingroup$ @JulianAngussmith If you think that I've answered your question, then you should mark my answer as the accepted one. If you have some other question, then post it as such. $\endgroup$ – José Carlos Santos Oct 23 '18 at 8:55
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    $\begingroup$ @user376343 Thanks. This post was helpful. I understand how the sum of the roots where calculated and the difference in the rational/integer root theorems. I still require further help in showing $p(z)$ has a single root in the first quadrant, but this will be asked in following posts. $\endgroup$ – JulianAngussmith Oct 23 '18 at 9:26

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