Show that $$P(z)=z^4+z^3+4z^2+2z+3$$ has no root in the first quadrant of the complex plane.

I've tried to do the exercise using argument principle, but I get the result that the polynomial has one root in the first quadrant.

  • 2
    $\begingroup$ And what is the root you found? Or, I would like to see how your argument works. Thanks. $\endgroup$ – awllower Jul 9 '13 at 5:54
  • $\begingroup$ The argument principle seems a good idea to me: start to select the cases $z$ real and $z$ purely imaginary before moving to the interior of the 1st quadrant: could you please summarize your computations? $\endgroup$ – Avitus Jul 9 '13 at 8:07

Applying the Routh-Hurwitz stability criterion to $P(z)$, we have to verify the conditions: i) all five coefficients are strictly positive (yes) ; ii) $a_3a_2 > a_4a_1$ (1*4> 1*3); iii) $a_3a_2a_1>a_4 a_1^2+a_3^2a_0 \,(1*4*2>1*2^2+1^2*3)$. All these are true in your case.


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