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I'm trying to understand Laguerre's method for polynomial root finding. However, I have some difficulties to understand one sentence of the book Applied Computational Complex Analysis (vol. 1) by Peter Henrici. There it says:

The function $f$ is known as the Laguerre iteration function. By algebraic manipulation it can be put in the form $$f(z)=z-\frac{\nu g(z)}{g'(z)+\{(\nu - 1)^2[g'(z)]^2-\nu(\nu-1)g(z)g''(z)\}^{1/2}}$$ where the argument of the root is to be chosen to differ by less than $\pi/2$ from the argument of $(\nu-1)g'(z)$.

Please, can anyone explain to me what does the last sentence mean? For starters, I assume that by "argument of the root" it means the sign in front of the square root, correct? But what is the "the argument of $(\nu-1)g'(z)$"?

How can anything be "less than" in complex arithmetic?

Also, I assume the purpose of this is to make the right term as small as possible. Why not simply take the larger of: $$g'(z)+\{(\nu - 1)^2[g'(z)]^2-\nu(\nu-1)g(z)g''(z)\}^{1/2}$$ and $$g'(z)-\{(\nu - 1)^2[g'(z)]^2-\nu(\nu-1)g(z)g''(z)\}^{1/2}$$ In other words, I would greatly appreciate if someone could translate that last line into something akin to:

if ? then
    sign = '+'
else
    sign = '-'

Thanks in advance!

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Any complex number can be written in the form $re^{i\theta}$, where $r \ge 0$, and $\theta$ is determined up to multiples of $2\pi$. Then we define the argument of $re^{i\theta}$ to be $\theta$.

http://en.wikipedia.org/wiki/Argument_%28complex_analysis%29

So suppose $z = r e^{i\theta}$. Then $\sqrt{z}$ is either $\sqrt{r} e^{i\frac12 \theta}$, or $\sqrt{r} e^{i(\frac12 \theta + \pi)}$, which have arguments $\frac12\theta$ and $\frac12 \theta + \pi$ respectfully. Given any other complex number $w = se^{i\psi}$, only one of the angles $\frac12\theta$ and $\frac12 \theta + \pi$ can be within $\frac\pi2$ of the angle $\psi$ (unless $\psi$ is exactly $\frac\pi2$ from $\frac12\theta$, in which case you can use either of them).

Your characterization of how to pick the sign is correct if $\nu > 1$. If $\nu<1$, then the exact opposite will be true (that is, pick the smaller of the two).

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  • $\begingroup$ Thanks a lot! $\nu$ apparently equals the degree of polynomial, so yes, it's $> 1$. Could you please explain to me, why would someone use the above method from the book as opposed to simply taking the larger denominator? Is it because of the pedagogical purpose? $\endgroup$ – Ecir Hana Sep 18 '14 at 14:57
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    $\begingroup$ I have no idea why they wrote it so awkwardly. $\endgroup$ – Stephen Montgomery-Smith Sep 18 '14 at 21:11

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