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Let $f(z) = z^n + a_{n-1}z^{n-1} + ... $ be a monic polynomial of degree $n$, and assume that $f(iy) \ne 0$ for all $y \in \mathbb R$. Write $f(iy) = u(y) + iv(y)$ and express the number of roots of $f$ in the right half plane {$\Re(z)> 0$} in terms of the number and mutual position of the real roots of $u$ and $v$.

Now we have the first constant is equal to 1 and I searched and got the idea that it is always possible to write $f(z)=u(x,y)+iv(x,y)$ in here. But I do not know where to start in this question. Any help is appreciated!

Cem Sarıer
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    Take inspirations from the answers in https://math.stackexchange.com/q/2258600/115115, https://math.stackexchange.com/q/2402864/115115, https://math.stackexchange.com/q/31651/115115, https://math.stackexchange.com/q/2472951/115115, https://math.stackexchange.com/q/2333348/115115 – Lutz Lehmann Jan 01 '18 at 14:08
  • @LutzL I couldn't write f(iy). I think I am missing some kind of simplification. – Cem Sarıer Jan 05 '18 at 19:53
  • What do you get if you literally substitute the variable $z$ for $iy$ in the formula for $f(z)$? Note that in general, for any $k \in \mathbb{N}$ and numbers $z_1$ and $z_2$, the equality $(z_1z_2)^k=z_1^kz_2^k$ holds --- and so in particular, $(iy)^k=i^ky^k$. (This is equal to either $y^k$, $iy^k$, $-y^k$ or $-iy^k$, depending on the value of $k$.) Since $y$ itself is assumed to be real, you can then collect together the imaginary terms and the real terms in your expression for $f(z)$ with $z:=iy$. – Julian Newman Jan 08 '18 at 14:37
  • I realise I'm assuming here that the coefficients $a_k$ are real - I guess this is meant to be the case in the question? But even if not, one can write $a_k=\alpha_k+i\beta_k$ for real $\alpha_k$ and $\beta_k$ and just get a more fiddly expression for $f(iy)$. – Julian Newman Jan 09 '18 at 03:14

1 Answers1

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You may look at the argument of $f(z)$ as $z$ goes along the boundary of a large half disk in the right half plane.

Thus you could consider the two contours:

$$z=R e^{i\theta}, -\pi/2\leq \theta\leq \pi/2$$ and $$ z=-it , \ -R\leq t \leq R .$$ For large enough $R$ the argument of $f(z)$ along the first contour will increase with $\pi n$.

The intertwined ordering of the real zeros (counting multiplicity) of $u$ and $v$ will tell you how the argument of $f(z)$ changes as you go along the imaginary axis. For example a zero of $v$ in between two zeros of $u$ tell you that the argument has increased (or decreased) by $\pi$. But for two consecutive zeros of $u$ there is no argument change but a reversal of the orientation.

It seems to me, however, that there is an information missing. For example, the polynomials $z^2\pm z+1$ will have the same locations for the real zeros of $(u,v) = (-y^2+1,\pm y)$ but two roots in the right half plane in one case and 0 in the other. The point is that you need to know the argument change at some point e.g. by knowing the signs of both $u$ and $v$ at, say $y=R$, but the given information only allows you to calculate one of them. Consequently an argument computation based on the location of real roots of $u$ and $v$ will give you either the number of roots in the right or the left half plane, but you can not tell which.

H. H. Rugh
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