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How many roots does $x^5+5x^3+2x^2+4x+1=0$ on the right half complex plane?

I notice that there can't be positive real roots, so we just need to count the root on the first quadrant and then multiply by 2. We have learnt Rouche's theorem and argument principle, but I have no clue about how to apply them. Thanks.

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This follows from the argument principle. We integrate over the $D$ shaped curve which starts from the positive imaginary axis say $iA$, down the imaginary axis to $-iA$ and then by a semicircle back to $iA$. We let $A\to \infty$. The difficult part is to find the image of this curve under the function $$f(x)=x^5+5x^3+2x^2+4x+1.$$ And in particular the number of times this image goes around the origin will be the number of roots. The semicircle part of the curve crosses the positive real axis $3$ times in the counterclockwise direction. For the segment along the imaginary axis we find

$$f(iy)=(y^5-5y^3+4y)i-2y^2+1$$

Thus this crosses the real axis only when $y=-2,-1,0,+1,+2$ and we see also that it only crosses the positive real axis at $y=0$. By looking at the graph of the imaginary part we see that at $y=0$ it is going in the clockwise direction so the total number of roots is $$3-1=2$$ Which coincides with the answer of Robert above (thanks, Robert.)

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  • $\begingroup$ Thanks, following your analysis and working out the details is quite fun. $\endgroup$ – A. Chu Apr 30 '17 at 5:59
  • $\begingroup$ Yes, I like these problems. $\endgroup$ – Rene Schipperus Apr 30 '17 at 6:00
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The roots are approximately $-.261532948669949$, $-.131757929232144 \pm .931292883370354 i$, $+.262524403567119 \pm 2.06232211872625 i$, so two in the right half plane.

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