the number of roots of a polynomial in the first quadrant

Feb 2011
147
19
I have to find out how many roots the polynomial

\(\displaystyle f(z)=z^8+3z^7-z^5+2z^4+z+1,\,z\in\mathbb{C}\)

has in the first quadrant.

Certainly, I need to use the argument principle, but I don't see how. Here's what I tried.

All zeroes of \(\displaystyle f\) lie in \(\displaystyle \{z\in\mathbb{C}:|z|<4\},\) which follows from applying the Cauchy bound. Suppose I can prove that \(\displaystyle f\) has no real and no imaginary roots. Then, since the function is holomorphic, I have to calculate

\(\displaystyle \frac{1}{2\pi i}\left (\int_{I_1}\frac{f'}{f}dz+\int_{I_2}\frac{f'}{f}dz+\int_{D}\frac{f'}{f}dz\right ),\)

where

\(\displaystyle I_1=\{z:0\leq Re(z)\leq 4,\,Im(z)=0\}\)
\(\displaystyle I_2=\{z:0\leq Im(z)\leq 4,\,Re(z)=0\}\)
\(\displaystyle D=\{z:0\leq Im(z),\,0\leq Re(z),\,(Re(z))^2+(Im(z))^2=16\},\)

(integrals have to be calculated in the appropriate directions of course).

But to calculate those integrals, don't I need to know the zeroes of \(\displaystyle f?\) Only then would I be able to integrate by residues, right? It's supposed to be very simple.
 

Opalg

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I have to find out how many roots the polynomial

\(\displaystyle f(z)=z^8+3z^7-z^5+2z^4+z+1,\,z\in\mathbb{C}\)

has in the first quadrant.

Certainly, I need to use the argument principle, but I don't see how. Here's what I tried.

All zeroes of \(\displaystyle f\) lie in \(\displaystyle \{z\in\mathbb{C}:|z|<4\},\) which follows from applying the Cauchy bound. Suppose I can prove that \(\displaystyle f\) has no real and no imaginary roots. Then, since the function is holomorphic, I have to calculate

\(\displaystyle \frac{1}{2\pi i}\left (\int_{I_1}\frac{f'}{f}dz+\int_{I_2}\frac{f'}{f}dz+\int_{D}\frac{f'}{f}dz\right ),\)

where

\(\displaystyle I_1=\{z:0\leq Re(z)\leq 4,\,Im(z)=0\}\)
\(\displaystyle I_2=\{z:0\leq Im(z)\leq 4,\,Re(z)=0\}\)
\(\displaystyle D=\{z:0\leq Im(z),\,0\leq Re(z),\,(Re(z))^2+(Im(z))^2=16\},\)

(integrals have to be calculated in the appropriate directions of course).

But to calculate those integrals, don't I need to know the zeroes of \(\displaystyle f?\) Only then would I be able to integrate by residues, right? It's supposed to be very simple.
For a start, f(z) has no positive real roots (it has the value 1 at z=0, and thereafter it increases, as you can check by elementary calculus). The polynomial has real coefficients, so its non-real roots occur in complex conjugate pairs. Therefore you can look for the number of roots with positive real part, and divide that number by 2 to get the number in the first quadrant.

So rather than looking at the winding number round a quadrant, I would look for the winding number of f(z) round a D-shaped contour consisting of a semicircle in the right-hand half plane, going from –iR to +iR, followed by a straight line segment down the imaginary axis from +iR to -iR. Here, R can be any number greater than or equal to 4.

To work out the winding number of f(z) as z goes round this contour, I would think in purely geometric terms rather than trying to evaluate the integral of f'/f. As z goes round the semicircle, f(z) is dominated by the \(\displaystyle z^8\) term , so it goes 4 times round the origin. For the straight line portion of the contour, write z=iy. Then \(\displaystyle f(iy) = (y^8 +2y^4+1) -i(3y^7+y^5+y).\) The real part of this is clearly always positive, so f(iy) cannot encircle the origin at all. Thus the winding number remains at 4, and the number of roots in the first quadrant is 2.
 
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Feb 2011
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Thank you very much, Opalg.

By f(z) being dominated by \(\displaystyle z^8,\) do you mean like in Rouché's theorem?
 

Opalg

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Thank you very much, Opalg.

By f(z) being dominated by \(\displaystyle z^8,\) do you mean like in Rouché's theorem?
Yes. (Nod)
 
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Feb 2011
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OK, thank you. I understand it now except that, very embarrassingly, I don't know what elementary calculus tells us that the polynomial increases in positive numbers. (Headbang)
 

Drexel28

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OK, thank you. I understand it now except that, very embarrassingly, I don't know what elementary calculus tells us that the polynomial increases in positive numbers. (Headbang)
Take the derivative!
 
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Opalg

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OK, thank you. I understand it now except that, very embarrassingly, I don't know what elementary calculus tells us that the polynomial increases in positive numbers. (Headbang)
Come to think of it, you don't even need calculus. If |x|<1 then \(\displaystyle 2x^4>x^5\), and if |x|>1 then \(\displaystyle 3x^7>x^5\). So in both cases, the positive terms outweigh the one negative term.
 
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Feb 2011
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Take the derivative!
Well, I did but apparently I'm even more of an idiot than I thought... I didn't see why the derivative was positive for positive reals.
 

Drexel28

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Well, I did but apparently I'm even more of an idiot than I thought... I didn't see why the derivative was positive for positive reals.
Repeatedly apply the same trick! It's initially positive, and the derivatives positive--to prove this its initially positive and the derivatives positive--to prove this....
 
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Feb 2011
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Repeatedly apply the same trick! It's initially positive, and the derivatives positive--to prove this its initially positive and the derivatives positive--to prove this....
I see, thanks.

(Crying)