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I'm considering the following 2nd degree polynomial for the case where the roots are complex conjugate.

$ P(z) = z^2 + (f^2 + f q -2)z + (1 - f q) = (z - z_1) (z - z^*_1) $

where f and q are real positive parameters.

Question 1: Is there a geometric interpretation of parameter q? There is a simple geometric interpretation of parameter f. It is $|z_1 - 1| = f$. In words, it is the distance between the roots to the point 1 on the real axis. But I can't find/formulate a geometric interpretation for q.

Question 2: This is a little bit more difficult to formulate but it is based on an observation I will explain. If parameters f and q are constrained to values in the set V = {0, 1, 2, ..., 2^N-1} / 2^(N-1). In words, f and q obtain values lying on equal interval grid points between 0 and 2.

If the possible roots of P(z) are plotted when f is fixed to any value in V and q is traced through V the roots lie on a circle with center (1,0) and radius f.

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If the possible roots of P(z) are plotted when q is fixed to any value in V and f is traced through V the roots lie on curves looking like this

enter image description here

If many more roots are plotted the pattern looks like this

enter image description here

It looks like the roots must lie on straight lines radiating from (1,0). I can't find any expression from which it is clear that this is true. So my questions is: Is it possible to find an expression that shows that the roots must line on straight line passing through (1,0)?

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  • $\begingroup$ There is a simple geometric interpretation of parameter f. No, $f = |z_1 - 1|$ is wrong. What's true is that $f^2 = P(-1)$, and also $|z_1 - 1|^2 = f^2 / (1 - f q)$, but neither has a direct geometric interpretation in terms of $f, q$ individually. $\endgroup$ – dxiv Jan 29 '16 at 0:51
  • $\begingroup$ hmm, are you sure? I mean, if $z_1 = Re^{i\theta}$ isn't $|z_1 - 1|^2 = R^2\sin^2(\theta) + (1 - R\cos(\theta) )^2 = 1 - 2R\cos(\theta) + R^2 = f^2$. Shouldn't it be $P(1) = f^2$ ? $\endgroup$ – niaren Jan 29 '16 at 7:28
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    $\begingroup$ You are right, it should have been $P(1) = f^2$. This follows from $Q(z) = P(z + 1) = f^2 + (f^2 - f q ) z + (1 - f q) z^2$ where obviously $Q(0) = f^2 = P(1)$. From the same $Q(z)$ however, it also follows that the product of the roots - which is the square of the absolute value for complex conjugates - is $f^2 / (1 - f q)$, where root $z_1$ of $Q(z)$ corresponds to root $z_1 - 1$ of $P(x)$. P.S. Note that your initial representation of $P(z)$ is off by a factor of $(1 - f q)$. It should be $P(z) = (1 - f q) (z - z_1) (z - z_1^*) $. $\endgroup$ – dxiv Jan 29 '16 at 7:54
  • $\begingroup$ Doh! $P(z)$ is not correct. Sorry! It was your comment regarding the factor $(1-fq)$ that made me see I wrote P(z) incorrectly.Sorry sorry sorry! I have update P(z) in the question. $\endgroup$ – niaren Jan 29 '16 at 8:08
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Let $P(z) = z^2 + (f^2 + f q − 2) z + (1 − f q)$ with $f, q \in \mathbb R$. The following will prove that if $f \ne 0$ and $|f + q| \lt 2$ then $P(z) = 0$ has two complex roots, and for each such root:

$$ |z - 1| = |f| \\ \cos \varphi = - sgn(f) \frac{f + q}{2} \,\,\,\,\text{ where }\,\, \varphi = \arg (z - 1)\,\,\text{ and }\,\, sgn \,\,\text{ is the signum function } $$

In the complex plane, this translates to:

  • for $f = const$ the locus of the roots $z$ of $P(z)$ is a circle centered at point $(1, 0)$;
  • for $f + q = const$ the locus of $z$ is a pair of straight lines through point $(1, 0)$.

Since the point of interest is $(1, 0)$ it is natural to consider $P(z) = Q(z - 1)$ around the origin. After simple calculations:

$$ \begin{align} Q(z) = P(z + 1) & = (z + 1)^2 + (f^2 + f q − 2) (z + 1) + (1 - f q) \\ & = z^2 + f (f + q) z + f^2 \end{align} $$

Its discriminant is:

$$ \begin{align} \Delta & = f^2 (f + q)^2 - 4 f^2 \\ & = f^2 (f + q - 2) (f + q + 2) \end{align} $$

Therefore the conditions for $\Delta \lt 0$ so that $Q(z)$ has complex roots are:

$$ \begin{cases} f \ne 0 \\ -2 \lt (f + q) \lt 2 \text{ } \iff \text { } |f + q| \lt 2 \end{cases} $$

Since the product of the two complex conjugate roots is $z \bar z = |z|^2$, it follows from Vieta's formulas that $|z|^2 = f^2$ which proves the first point.

Then the two roots are $z, \bar z = |f| ( \cos \varphi \pm i \sin \varphi)$ and their sum is $z + \bar z = 2 |f| \cos \varphi = -f (f + q)$ again by Vieta's formulas. It finally follows that:

$$ \cos \varphi = - \frac{f}{|f|} \frac{f + q}{2} = - sgn(f) \frac{f + q}{2} $$


[EDIT] This is a followup to the question of a geometric interpretation of q asked in the original post, and again in a comment. Short answer is that there doesn't appear to be any direct "nice" geometric interpretation of $q$, and here is why.

The answer above focused on the roots of $Q(z)$, and the geometric interpretations for $f$ and $f+q$ followed as corrolaries, almost as if by chance. However, the problem can be approached from the other end, too. Suppose one wants to directly determine how the parameters affect the roots, specifically what the locus is of roots $z, \bar z$ if a parameter is kept constant. Below is a derivation of the answer for $f$ since that's the easier of the two to calculate, and also provides the basis of why there is no comparably simple geometric interpretation for $q$.

The two complex conjugate roots $z, \bar z$ of$Q(z)$ satisfy the simultaneous equations.

$$ \begin{cases} z^2 + f (f + q) z + f^2 = 0 \\ {\bar z}^2 + f (f + q) {\bar z} + f^2 = 0 \end{cases} $$

$q$ can be easily eliminated between the two equations - by solving the first one in $q$ then substituting in the second one, or simply by multiplying the two equations by $\bar z, z$ respectively, then subtracting. In the end it reduces to:

$$ (z - \bar z) (|z|^2 - f^2) = 0 $$

Since the roots are non-real complex numbers $z \ne \bar z$, so it follows that $|z|^2 - f^2 = 0$. This is the same relation derived in the original answer, and has a nice direct geometric interpretation. It means that $|f|$ is the absolute value of $z$ so the roots for $f = const$ lie on a circle centered at the origin.

It is technically possible to do the same analysis for $q$ of course. But since the equations are quadratic in $f$, eliminating it between the two is more laborious, and the end results is a quartic curve parameterized in $f$. Each $q = const$ gives a different quartic curve, but other than that there is no obvious (to me) geometric interpretation for $q$.

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  • $\begingroup$ I'm going to accept this answer. Particularly, I like the idea of analyzing the modified polynomial $Q(z)$. Unfortunately, this idea would never have occurred to me. I think I derived an alternative expression for $\cos \varphi$. This is something I will look into. So the reason for the lines radiating from (1,0) is because $\varphi$ depends 'only' on $f+q$...It would be interesting if you could give your formulation of a geometric interpretation of $q$... $\endgroup$ – niaren Jan 31 '16 at 21:29
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    $\begingroup$ @niaren I don't think there really is much of a geometric interpretation of q - see the Edit at the end of my answer. $\endgroup$ – dxiv Jan 31 '16 at 23:46
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    $\begingroup$ thanks for the edit. The alternative expression I mentioned above was for the slope of the lines and it also depends on $f+q$ only. The reason I thought that there ought to be some 'nice' geometric interpretation/understanding of $q$ is because of $f q = 1 - R^2$ where $R$ is the modulus of the roots of $P(z)$. So it looks a bit like pythagoras or some simple trigonometric identity could be used to obtain som geometric understanding.. $\endgroup$ – niaren Feb 2 '16 at 9:26

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