# roots of $f(z)=z^4+8z^3+3z^2+8z+3=0$ in the right half plane

This is a question in Ahlfors in the section on the argument principle: How many roots of the equation $f(z)=z^4+8z^3+3z^2+8z+3=0$ lie in the right half plane?

He gives a hint that we should "sketch the image of the imaginary axis and apply the argument principle to a large half disk."

Since $f$ is an entire function, I think I understand that the argument principle tells us that for any closed curve $\gamma$ in $\mathbb{C}$, the winding number of $f(\gamma)$ around 0 is equal to the number of zeros of $f$ contained inside $\gamma$.

How would you go about actually applying the hint though? I am having trouble figuring out what the image of a large half disk under $f$ would look like.

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Some analysis of the behavior on the imaginary axis allows you to tell the net change in argument over the diameter of the semicircle, from $Ri$ to $-Ri$ for some big positive $R$. Noticing that the real part of $f(it)$ is $t^4-3t^2+3=\left(t^2-\frac{3}{2}\right)^2+\frac{3}{4}>0$, you can see first of all that there can be no winding about zero on the imaginary axis. Furthermore, since the real part of $f$ on the imaginary axis has degree $4$ and the imaginary part of $f$ on the imaginary axis has degree $3$, the real part of $f(\pm Ri)$ will be much larger than the imaginary part of $f(\pm Ri)$ for large $R$, which means that the argument of $f$ at the endpoints of the diameter will be near zero, and you can conclude that there is near $0$ net change in argument along the diameter of the semicircle.

This leaves the analysis of the change in argument of $f(Re^{i\theta})$ as $\theta$ ranges from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. For this, it is helpful to note that $\frac{1}{R^4}f(Re^{i\theta})=e^{4i\theta}\left(1+\frac{8}{R}e^{-i\theta}+\frac{3}{R^2}e^{-2i\theta}+\frac{8}{R^3}e^{-3i\theta}+\frac{3}{R^4}e^{-4i\theta}\right)$ is near $e^{4i\theta}$ when $R$ is large.

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Wow, that estimate for the semicircle is really neat! Thanks! So that tells you that the change in argument is $4\pi$ right? – user1736 Apr 8 '11 at 6:55
@user1736: Yeah, because you can get it as close to $4\pi$ as you want by taking $R$ big enough. – Jonas Meyer Apr 8 '11 at 12:28

The hint was to sketch the image of the imaginary axis. You can parametrize the imaginary axis as $\gamma(t) = it$. $f(it) = t^4-3t^2+3 + (-8t^3 + 8t)i.$ Do you recall how to plot the parametrized curve $(t^4-3t^2+3,-8t^3+8t)$ from calculus?

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in this case $\mathrm{Re} f(it)=t^4-3t^2+3$ has no roots, so $f(it)$ is alwas in the right half-plane, and for $t\to \pm\infty$ it "goes in the direction of the $x$-axis" (as the imaginary part of $f(it)$ grows more slowly). Hence $\mathrm{Arg} f(it)$ makes $0$ turns as $t$ goes from $-\infty$ to $+\infty$

If you now take $(-iN,iN)$ for a large $N$, and complete it to a closed curve by a semicircle on the right, the number of turns of $\mathrm{Arg} f(it)$ comes only from the semicircle, and so it's $4/2$ ($4$ is the degree of $f$) $=2$.

Your polynomial has $2$ roots in the right half-plane.

In more general situation, one needs to find the total number of turns of $\mathrm{Arg} f(it)$ (it is a half-integer if deg of $f$ is odd). The method is to find the intersection of the curve $f(it)$ with the real and the imaginary axis, to see how it enters and leaves the quadrants of the plane. So one needs to find the real roots of $\mathrm{Re} f(it)$ and of $\mathrm{Im} f(it)$ - in fact, only their relative positions on the real axis.

It is actually quite useful - e.g. if $f$ is the characteristic polynomial a system of linear ord. diff. equations with constant coefficients, then no roots in the right half-plane = stability of the system.

edit: More about the method of finding the number of roots with positive real part. Let $f(z)=z^n+a_{n-1}z^{n-1}+\dots+a_0$, $a_i$'s are complex numbers. Let us decompose $f(it)$ as $f(it)=p(t)+iq(t)$, where $p$ and $q$ are real polynomials.

First we identify in which quadrant $(p(t),q(t))$ is when $t\to -\infty$ (its given just by the signs of the leading coefficients of $p$ and $q$). Then we draw on the real axis the real roots of $p$ and of $q$ (and mark multiplicities). Suppose that $p$ and $q$ have no common real root, i.e. that there is no purely imaginary root of $f$ (otherwise we need to shift $z$ by a small $\epsilon$). We don't need the exact values of the roots of $p$ and $q$; only their relative positions on the real axis is used.

Now proceed on the real axis from $t=-\infty$ to $+\infty$. Whenever you meet a root of either $p$ or of $q$, you change the quadrant. In fact, if you meet two roots of $p$ without a root of $q$ in between them, it means that you return to your original quadrant. So simply erase such pairs of roots until between any roots of $p$ (or $q$) there's a root of $q$ (or $p$). Now every remaining root means a change of the quadrant and those changes go in the same direction (clockwise or anticlockwise). So $1$ plus the number of the remaining roots, divided by $4$, is the number of turns of $f(it)$ as $t$ goes from $-\infty$ to $\infty$. (divided by $4$ as we count quadrants).

If we take a large right half-circle, its $f$-image makes $n/2$ turns. If we subtract from this $n/2$ the number we found above (number of turns of $f(it)$), we get the number of roots with positive real part.

Also notice that $\deg p=n$, $\deg q\leq n-1$, so the (absolute value of the) number of turns of $f(it)$ is at most $(1+n+(n-1))/4=n/2$. So if you want to have no root with positive real part (or no root with negative real part) then $p$ and $q$ must have all roots real and they must alternate.

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Sorry, but could you elaborate on your second paragraph? How do you know that the completion of the closed curve by a semicircle on the right circles around the origin? – user1736 Apr 8 '11 at 6:51
@user1736: it's just $f(Ne^{i\alpha})=N^n e^{in\alpha}(1+\text{something small})$, where $n$ is the degree of $f$ - so it makes $n/2$ turns. – user8268 Apr 8 '11 at 8:50
@user8268, so thats basically what Jonas was talking about at the end of his response right? – user1736 Apr 11 '11 at 15:30
@user1736: yes, my answer is simply identical to his. The only added value is the method described after editing (and the fact that it's actually useful:) (used in engineering) – user8268 Apr 11 '11 at 19:54