Sign of Laguerre root finding iteration

Question

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!

Answer 1

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).