Calculate the number of zeros in the right half-plane of the following polynomial: $$z^4+2z^3-2z+10$$ Please, it's the last exercise that I have to do. Help TT. PD: I don't know how do it.

  • $\begingroup$ By the "right half-plane", do you mean $x+iy \in \mathbb{C}$, with $y>0$ or $y \ge 0$?! $\endgroup$ Feb 17, 2013 at 2:09
  • $\begingroup$ I mean $Re(z)>0$ $\endgroup$ Feb 17, 2013 at 2:14
  • 2
    $\begingroup$ What's "it's the last exercise that I have to do" supposed to mean? If you don't master a particular area, you should do more exercises to practice, no matter what "list" you're working off. $\endgroup$
    – mrf
    Feb 17, 2013 at 9:12
  • $\begingroup$ I suppose using the formula for the roots of quartic polynomials is cheating? $\endgroup$
    – vonbrand
    Feb 17, 2013 at 16:37

2 Answers 2


Proceed like the previous problem for first quadrant. You will find one root. And note that roots will be conjugates. So 2 roots in the right half-plane..

For zero in the first quadrant, consider the argument principle: if $Z$ is the number of zeroes of $f$ inside the plane region delimited by the contour $\gamma$, then $\Delta_\gamma(\textrm{arg}f)=2\pi Z$, i.e. the variation of the argument of $f$ along $\gamma$ equals $Z$ times $2\pi$.

Take a path from the origin, following the real axis to the point $M>0$, then make a quarter of circle or radius $M$, reaching the point $iM$ and then go back to the origin along the imaginary axis. Now try to determine the variation of the argument of $f(z)$ along this path for $M\to\infty$:

  1. along the real axis, the function is $f(t)=t^4-2t+2t^3+10$, therefore $f(t)$ is real for $t\geq0$ so the total change of argument along this part of the path is $0$.
  2. along the path $Me^{i\theta}$ for $0\leq\theta\leq \pi/2$, if $M$ is very large, the function is near to $g(\theta)=M^4e^{i4\theta}$; therefore the argument goes from $0$ to $2\pi$.
  3. along the imaginary axis, the function's argument doesn't change. So, the total change of the argument is $2\pi$, implying that the function has only one zero in that quadrant.
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    $\begingroup$ I don't understand. "Proceed like the previous problem for first quadrant"? $\endgroup$ Feb 17, 2013 at 2:55
  • $\begingroup$ Use the " argument principle " as in [math.stackexchange.com/questions/305658/… $\endgroup$
    – Halil Duru
    Feb 17, 2013 at 3:04
  • $\begingroup$ sorry, but I do not serve your explication. $\endgroup$ Feb 17, 2013 at 3:08
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    $\begingroup$ But the argument of the polynomial does change on the imaginary axis. Here's the image of your path when M=2 and when M=20. $\endgroup$ Feb 17, 2013 at 10:02
  • $\begingroup$ As you see the change in the argument along the imaginary axis approaches $0$ as $M$ approaches $\infty$, right? $\endgroup$
    – Halil Duru
    Feb 17, 2013 at 12:10

Consider the polynomial

$$ p(z) = z^4 + 2z^3 - 2z + \epsilon. $$

When $\epsilon = 0$ we can solve for the zeros explicitly using the cubic formula. They are $z=0$ and

$$ z \in \left\{\begin{array}{c} -\frac{2}{3}+\frac{1}{3} \sqrt[3]{19-3 \sqrt{33}}+\frac{1}{3} \sqrt[3]{19+3 \sqrt{33}}, \\ -\frac{2}{3}-\frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{19-3 \sqrt{33}}-\frac{1}{6} \left(1-i \sqrt{3}\right) \sqrt[3]{19+3 \sqrt{33}}, \\ -\frac{2}{3}-\frac{1}{6} \left(1-i \sqrt{3}\right) \sqrt[3]{19-3 \sqrt{33}}-\frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{19+3 \sqrt{33}} \end{array}\right\}. $$


$$ \sqrt[3]{19-3 \sqrt{33}} > \sqrt[3]{19-3 \sqrt{36}} = 1, $$

so that the first zero in this set satisfies

$$ \begin{align*} & -\frac{2}{3}+\frac{1}{3} \sqrt[3]{19-3 \sqrt{33}}+\frac{1}{3} \sqrt[3]{19+3 \sqrt{33}} \\ &\qquad > -\frac{2}{3}+\frac{1}{3} \sqrt[3]{19-3 \sqrt{33}}+\frac{1}{3} \sqrt[3]{19-3 \sqrt{33}} \\ &\qquad = -\frac{2}{3}+\frac{2}{3} \sqrt[3]{19-3 \sqrt{33}} \\ &\qquad > 0. \end{align*} $$

The real part of the next two zeros is clearly negative:

$$ -\frac{2}{3}-\frac{1}{6} \sqrt[3]{19-3 \sqrt{33}}-\frac{1}{6} \sqrt[3]{19+3 \sqrt{33}} < 0. $$

So when $\epsilon = 0$ we have one zero at $z=0$, one zero with $\Re(z) > 0$, and two zeros with $\Re(z) < 0$.


$$ p(iy) = y^4 + \epsilon - i2(y+y^3), $$

so that $p(z)$ has no zeros on the imaginary axis when $\epsilon > 0$.

We now consider the zero of $p(z)$ which is located at $z=0$ when $\epsilon = 0$ as an analytic function $z_0 = z_0(\epsilon)$ with $z_0(0) = 0$. Expand $z_0$ as a Taylor series

$$ \begin{align*} z_0(\epsilon) &= z_0(0) + z_0'(0)\epsilon + O(\epsilon^2) \\ &= z_0'(0)\epsilon + O(\epsilon^2) \end{align*} $$

(valid for small $\epsilon$) and substitute this into the equation $p(z_0) = 0$ to get

$$ [1-2z_0'(0)]\epsilon = O(\epsilon^2). $$

Divide both sides by $\epsilon$ and let $\epsilon \to 0$ to find that

$$ 1-2z_0'(0) = 0 $$

or $z_0'(0) = 1/2$. Thus for small $\epsilon > 0$ there are two zeros satisfying $\Re(z) > 0$. Since $p(z)$ has no zeros with $\Re(z) = 0$ when $\epsilon > 0$ and the zeros of polynomials are continuous functions of the coefficients we conclude that this is true for all $\epsilon > 0$.

In particular, when $\epsilon = 10$ there are exactly two zeros with $\Re(z) > 0$ and two zeros with $\Re(z) < 0$.


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