I have a polynomial $f(z)=z^4+z^3-2z^2+2z+4$, and I want to find the number of roots in the first quadrant. I'm trying to use the argument principle (or Rouche), and I could try to make my contour the quarter circle, but I've having trouble because I can't justify that there are no roots on the real axis. Please give me some recommendations!

So now I understand why there are no roots on the contour; I have also justified that the integral on the arc goes to $2\pi i$ by normal limit considerations. However, I still am unsure how to figure out the arguments.

  • $\begingroup$ $-1$ is a root so one can factor it $\endgroup$ – Conrad May 22 at 18:18
  • $\begingroup$ $z=-1$ is an obvious root. So factor out $(z+1)$ and then you get a cubic polynomial on which you can try Rouche's theorem. $\endgroup$ – Anurag A May 22 at 18:19
  • $\begingroup$ @Conrad But I understand I could just figure out that $-1$ and $-2$ are roots and factor the remaining part, but I'd like to try to solve this explicitly with the theorems I mentioned above. $\endgroup$ – user790925 May 22 at 18:20
  • $\begingroup$ on the positive axis there are no roots because $z^4-2z^2+1 \ge 0$ and then you add to it only positive numbers when $z>0$ $\endgroup$ – Conrad May 22 at 18:23
  • $\begingroup$ @Conrad I see, that makes sense... however afterwards I'm still struggling to understand what the change in argument would be. $\endgroup$ – user790925 May 22 at 18:27

This one is a real weird one.

That is because I am tempted to factor this polynomial.

Let us try the rational root theorem first. If there is a rational root $\frac pq$ then $p$ divides $4$ and $q$ divides $1$ i.e. that root must be a multiple of $4$.

Merely trying out $z = -1$ works. Upon division by $z+1$ we get $z^3-2z+4$.

Another use of RRT gives $z = -2$ as a root, and division by $z+2$ yields $z^2-2z+2$, which by the usual quadratic formula is $(z-1+i)(z-1-i)$.

So, we have the roots as $1\pm i, -1,-2$. Of these, exactly the root $1+i$ is in the positive quadrant.

While this may be disappointing as an answer because we have not used machinery, it is suitable for a beginner. I would always suggest it as a first approach, and then look to apply Rouche or something else if things did not work out.

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You can use the Rouche method if you prove that there are no roots on the positive real axis, which is the part of the real axis that's on your path.

When $x>\sqrt2$ (I'm using $x$ here to emphasize we're dealing with a real variable in the proof), $x^4$ has a greater absolute value than $-2x^2$ and the latter is the only negative term, thus the polynomial is forced (strictly) positive. Similarly, $4$ dominates $-2x^2$ for $0<x<\sqrt2$ and the sum $x^4+4$ dominates $-2x^2$ for the remaining case $x=\sqrt2$.

So there must be no positive real roots and you can use a Rouche path that includes the positive real axis. You should, of course, compare your result with that obtained by elementary techniques from Aston.

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