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I am stuck on this problem:
a) let a be a complex number, M and M' the images of a and $\bar a$. Study the sign of Im{$\frac{1}{z-a}+\frac{1}{z-\bar a}$} with the position of the point P image of z d to the real axis and to the circle of diameter MM'.

b) Be A a polynomial in R[X]. Be $\rho_1$, $\bar \rho_1$, $\rho_2$, $\bar \rho_2$,..$\rho_p$, $\bar \rho_p$, the complex roots of A. Be $M_i$, $M'_i$ the images of $\rho_i$ and $\bar \rho_i$, $\Gamma_i$ the circle of diameter $M_i$$M'_i$. Show that all non real roots of the derived polynomial A' has its image either within of one of the $\Gamma_i$ or on one of the $\Gamma_i$.

For part a), I tried to express the Imaginary part of $\frac{1}{z-a}+\frac{1}{z-\bar a}$ by decomposing with the real and imaginary parts of a and z and using the fact that $\frac{1}{z}= \frac{\bar z}{|z\bar z|}$, but nothing convincing came up.
For part b) I think this has to do with the Gauss Lucas theorem. Thanks a lot for support

Matfi
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1 Answers1

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This is Jensen's theorem.

Let $p\in{}\mathbb{R}[x]$. It is known that if $a$ is a complex root of $p$, Then $\bar a$ is also a root. In this case, we will say that the disk of diameter $aa^*$ (i.e. its diameter is the line between $a$ and $\bar a$) is a Jensen Disk of $p$ (example: http://mathworld.wolfram.com/JensenDisk.html).

Theorem: all nonreal roots of $p'$ lie in or on the boundary of one the Jensen Disks of $p$.

To prove this, let $z_1,...,z_n$ be the roots of $p$. Assume that $z$ is a nonreal root of $p'$ that is not in or on the boundary of a Jensen Disk of $p$. Note that $z$ is not a root of $p$, and thus $0=\frac{p'(z)}{p(z)}=\sum_{i=1}^n{\frac{1}{z-z_i}}$.
We shall reach a contradiction by showing that the imaginary part of $\sum_{i=1}^n{\frac{1}{z-z_i}}$ is not zero. we will do it by showing that:
(i) $sgn(\Im(\frac{1}{z-a}+\frac{1}{z-\bar a}))=-sgn(\Im(z))$ for each complex nonreal root $a$ of $p$ (which is your question (a)), and that
(ii) $sgn(\Im(\frac{1}{z-r}))=-sgn(\Im(z))$ for each real root $r$ of $p$.

Indeed, this would imply that $sgn(\Im(\sum_{i=1}^n{\frac{1}{z-z_i}}))=-sgn(\Im(z))\ne{}0$.

i) Let $a=s+ti$ be a nonreal root of $p$. Then

$$\frac{1}{z-a}+\frac{1}{z-\bar a}=\frac{1}{z-s-ti}+\frac{1}{z-s+ti}=\frac{2z-2s}{(z-s)^2+t^2}=\frac{2(z-s)((\bar z-s)^2+t^2)}{|(z-s)^2+t^2|^2}$$

Define $x\sim{}y$ if $sgn(\Im(x))=sgn(\Im(y))$. So

$$ \frac{1}{z-a}+\frac{1}{z-\bar a}= \frac{2(z-s)((\bar z-s)^2+t^2)}{|(z-s)^2+t^2|^2}\sim{} (z-s)((\bar z-s)^2+t^2)=$$ $$=|z-s|^2(\bar z-s)+(z-s)t^2\sim{} |z-s|^2\cdot(-z)+z\cdot{}t^2= (t^2-|z-s|^2)z$$

But the expression $t^2-|z-s|^2$ is negative since $z$ is not in the Jensen disk of diameter $a \bar a$. Thus $sgn(\Im(\frac{1}{z-a}+\frac{1}{z-\bar a}))=-sgn(\Im(z))$, as required.

ii). Let $r$ be a real root of $p$. Then

$$ \frac{1}{z-r}-\overline{\frac{1}{z - r}}= \frac{1}{z-r}-\frac{1}{\bar z - r}= \frac{\bar z - z}{|z-r|^2}\sim{}\bar z - z\sim{} -z$$

And it follows that $sgn(\Im(\frac{1}{z-r}))=-sgn(\Im(z))$, as required.

Thus, we have reached the desired contradiction.

idok
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