I want to find the roots of $$f(z)=\left[a+zg(z)\right]^2+g(z)^2=0$$
Where $a$ is real number and: $$ g(z)=\frac{1}{2\sqrt{z^2+1}}\ln\left(\frac{z+\sqrt{z^2+1}}{z-\sqrt{z^2+1}}\right) $$
It is known that $f(z)=0$ has double complex roots when $a\in(-\pi/2,\pi/2)$, no roots when $a>\pi/2$ and four complex roots when $a<-\pi/2$. It also says that the complex roots are purely imaginary and comes in conjugate pairs. (The conclusion can be visualized by the accepted answer here.)
For example:
- when $a=-\pi$, $z=\pm 8.02398 i, \quad\pm14.5019 i$.
- when $a=-\pi/6$, $z=\pm 1.62943 i $
- when $a=\pi/6$, $z=\pm 0.556395 i $
Note if you have an analytical answer in mind for the above conclusion you can stop here and kindly post it in an answer.
I want a more analytical method to get that conclusion, so I did this: $$ a=-\frac{z\pm\mathrm{i}}{2\sqrt{z^2+1}}\ln\left(\frac{z+\sqrt{z^2+1}}{z-\sqrt{z^2+1}}\right). $$ Let $z=\sinh q$, then $\sqrt{z^2+1}=\cosh q$. $$ \ln\left(\frac{z+\sqrt{z^2+1}}{z-\sqrt{z^2+1}}\right)=2q + 2n\pi+\pi i \quad n\in \mathcal{N} $$
So we get: $$ a=-\frac{\sinh q \pm i}{\cosh q}(q+n\pi i+ \frac{\pi}{2}i) $$
Let $p=q+n\pi i+ \frac{\pi}{2}i$, then $z=\sinh q = (-1)^{n+1}i\cosh p$, and $\cosh q = (-1)^{n+1} i \sinh p$
Substitute to the expression of $a$: $$ a=-\frac{\cosh p\pm 1}{\sinh p}p = -p\coth\frac{p}{2} \qquad\text{or} \qquad -p\tanh\frac{p}{2} $$
Recall that $a$ is real, so $p$ must be purely real or purely imaginary(Is this claim true?), but $z=\pm i \cosh p$ is always purely imaginary. So I've proved that the roots must be purely imaginary.
If $p=ip_0$ is purely imaginary, $z=\pm i \cos p_0$. The form of the equation for $a$ reads: $$ a = p_0 \cot \frac{p_0}{2} \quad\text{ or }\quad p_0 \tan \frac{p_0}{2}$$
First we reckon $p$ to be real, and plot out the RHS of w.r.t $p$:
When $p=-\pi$, we read out from the graph, $p=2.77168$ and $3.36624$. $z=\pm \cosh p$ give us the four desired root. But the problem is, when $a\in(-2,-\pi/2)$, there seems only one pair of roots from the graph.
When $p=-\pi/6$, we read out from the graph, that $p=1.07018$, $z=\pm i\cosh p=\pm 1.62943 i$, which again give us the desired pair of roots. However, what if $a>0$, for example $a=\pi/6$, can't read out from this graph.
Next we reckon $p=ip_0$:
There are many crossings, read the ones most close to the $y$ axis(I don't have a particular reason).
when $p=\pi/6$, $p_0 = 0.980755$ and $z=\pm i\cos p_0 = \pm 0.556395 i$. We get the desired results, but why should I choose the one nearest to the $y$ axis, there are many roots after all. Besides, it seems that from this graph that $a>\pi/2$ we still have solutions, which is not right. I also want to mention that, the intersection of blue and orange lines has the height of $\pi/2$.
when $p=-\pi/6$ or $-\pi$, we can never get the right results, for $|\cos p_0| <=1$.
Question is that what's the hidden flaws in this derivation? How can I fix them so that I can get the desired conclusion and results?
I don't quite get this: $$\ln\left(\frac{z+\sqrt{z^2+1}}{z-\sqrt{z^2+1}}\right)=2q + 2n\pi+\pi i \quad n\in \Bbb{N}$$ (which should be $n \in \Bbb Z$, anyway).
Since $z = \sinh q$, $$\begin{align} \ln\left(\frac{z+\sqrt{z^2+1}}{z-\sqrt{z^2+1}}\right) &= \ln\left(\frac{\sinh q+\cosh q}{\sinh q-\cosh q}\right)\\&=\ln\left(\frac{(\sinh q+\cosh q)^2}{-1}\right)\\&=\ln \left(-(e^q)^2\right)\\&=\ln e^{2q + \pi i}\\&=2q + \pi i + 2n\pi i = 2q + (2n+1)\pi i\qquad n\in \Bbb Z. \end{align}$$
Of course, I've also followed your lead in being incautious about $\sqrt{z^2+1} = \cosh q$.