Location of zeros of $x^{2} + 2z_{0}x + 1$ Suppose $z_{0}$ is a complex number with nonzero imaginary part. The polynomial $x^{2} + 2z_{0}x + 1$ has one zero in $|z| < 1$ and one zero in $|z| > 1$. The zeros are $-z_{0} \pm \sqrt{z_{0}^{2} - 1}$ where we have defined the square root of a complex number as follows: Suppose $z = re^{i\theta}$, $0 \leq \theta < 2\pi$. Then $\sqrt{z} := \sqrt{r}e^{i\theta/2}$.
When is $|-z_{0} + \sqrt{z_{0}^{2} - 1}| < 1$ and when is $|-z_{0} - \sqrt{z_{0}^{2} - 1}| < 1$? My guess is that the former occurs when $|z_{0}| > 1$ and the latter occurs when $|z_{0}| < 1$, but I don't know how to show this.
 A: It's not the modulus of $z_0$ that determines whether $\lvert -z_0 +\sqrt{z_0^2-1}\rvert < 1$. If you look at the imaginary axis,
$$\left\lvert -it + \sqrt{(it)^2-1}\right\rvert = \left\lvert -it + \sqrt{-(t^2+1)}\right\rvert = \left\lvert -it + i\sqrt{t^2+1}\right\rvert = \left\lvert \sqrt{t^2+1}-t\right\rvert$$
is smaller than $1$ if and only if $t > 0$.
Since the square root is always in the upper half-plane, one may conjecture that the criterion is whether $\operatorname{Im} z_0 > 0$ or $< 0$.
That is indeed the case.
Consider the map
$$\psi \colon w \mapsto \frac{1}{2}\left(w + \frac{1}{w}\right).$$
It is a rational function of order $2$ and satisfies $\psi(1/w) = \psi(w)$, so its restriction $\varphi$ to the upper half-plane is a biholomorphism between the upper half-plane and $\widehat{\mathbb{C}}\setminus \psi(\mathbb{R}\cup\{\infty\}) = \mathbb{C}\setminus \{x \in \mathbb{R} : \lvert x\rvert \geqslant 1\}$. Also,
$$2 \operatorname{Im} \psi(w) = \operatorname{Im} w + \operatorname{Im} \frac{1}{w} = \left(1 - \frac{1}{\lvert w\rvert^2}\right)\operatorname{Im} w,$$
so $\operatorname{Im}\psi(w) > 0$ if and only if a) $\operatorname{Im} w > 0$ and $\lvert w\rvert > 1$, or b) $\operatorname{Im} w < 0$ and $\lvert w\rvert < 1$.
Furthermore, when $z$ is in the upper half-plane, then so is $z+\sqrt{z^2-1}$ by the choice of the square root, and
$$\varphi(z+\sqrt{z^2-1}) = \frac{1}{2}\left((z+\sqrt{z^2-1}) + (z-\sqrt{z^2-1})\right) = z$$
also lies in the upper half-plane, which by the above implies $\left\lvert z+\sqrt{z^2-1}\right\rvert > 1$.
The case for the lower half-plane is similar, so we have
$$\left\lvert -z_0 + \sqrt{z_0^2-1}\right\rvert < 1 \iff \operatorname{Im} z_0 > 0.$$
A: A more visual approach might prove useful. If $$z^2+2z_0z+1=0,$$ then $$x^2-y^2+2(x_0x-y_0y)+1=0\ \ \ \ \ \text{and}\ \ \ \ \ 2xy+2(x_0y+y_0x)=0,$$ hence, $$(x+x_0)^2-(y+y_0)^2=x_0^2-y_0^2-1\ \ \ \ \ \text{and}\ \ \ \ \ y=-\frac{y_0x}{x+x_0},$$ a hyperbola and rational function. Thus, if we assume $x_0^2-y_0^2-1>0$, $x_0,y_0>0$ and define $R=\sqrt{x_0^2-y_0^2-1}$, then we arrive at something like, 
