A paper described a contour as following with figure given below: From $A$ to $B$, it is an ellipse with foci $\pm 1$ and $|z+\sqrt{z^2-1}| = 2 $, from $B$ to $C$ the line $z=1+re^{i \varphi}$, $\frac{3\pi}{2} \le \varphi \le 2\pi$, from $C$ to $D$ a circle with center $z = 1$ with radius $\rho$ and from $D$ to $A$, the line $z = 1+re^{i\varphi}$, $0 \le \varphi \le \frac{\pi}{2}$.
My doubt is that I don't understand how the contour is a line from $A$ to $D$ and $C$ to $D$. Also, the paper doesn't specify anything about $r$. What did the paper meant?
Edit 1: Direct snapshot from the paper:
Let $D$ be the point $1+\rho_D e^{i\varphi}$. Both $\rho_D$ (distance from $z_0=1$ to $D$) and $\varphi$ are fixed.
Let $A$ be the point $1+\rho_A e^{i\varphi}$. $\rho_A$ (distance from $z_0$ to $A$) is also fixed.
Now parameterize the directed line segment $\overline{DA}$ by a real variable $r$ with the function
$$z(r) = 1 + re^{i\varphi}$$
where $r\in[\rho_D,\rho_A]$.
I believe what the author is trying to communicate is that $\varphi$ is indeed fixed, but that it's chosen from the interval $\left[0,\frac\pi2\right]$ so that the point $D$ lies anywhere up and to the right of $z_0$.
The line segment $\overline{BC}$ can be parameterized in the same way, just fix a different $\varphi\in\left[\frac{3\pi}2,2\pi\right]$. This ensures that $B$ lies down and to the right of $z_0$. For the sake of symmetry, for whatever $\varphi$ you select for $\overline{DA}$, pick $2\pi-\varphi$ for $\overline{BC}$.
Addressing a comment:
That's how we've defined $A$, a point $\rho_A$ units away from $z_0$ along a ray $\varphi$. Although $\rho_A$ is constant, its exact value depends on the constraint imposed by the ellipse equation.
Given $\varphi$, one would then have to solve
$$\left|1 + \rho_A e^{i\varphi} + \sqrt{\left(1+\rho_A e^{i\varphi}\right)^2-1}\right| = R$$
for $\rho_A$ to determine the exact coordinates of $A$.
As an example, suppose we choose $\varphi=0$. Then $A:=1+\rho_A$ would lie on the real axis, and using the constraint enforced by the ellipse we find
$$\left|1 + \rho_A + \sqrt{\left(1+\rho_A\right)^2-1}\right| = R \implies \rho_A = \frac{(R-1)^2}{2R}$$
Any other choice of $\varphi\in\left(0,\frac\pi2\right]$ would make solving for $\rho_A$ quite a bit more complicated, but we can still do so systematically.
As another example, suppose we pick $\varphi=\frac\pi4$. Then
$$\begin{align*} \left|1 + \rho_A e^{i\pi/4} + \sqrt{\left(1+\rho_A e^{i\pi/4}\right)^2-1}\right| &= R \\ \left|1 + \frac{\rho_A}{\sqrt2} + i \frac{\rho_A}{\sqrt2} + \sqrt{\sqrt2\,\rho_A + i\left(\sqrt2\,\rho_A + \rho_A^2\right)}\right| &= R \end{align*}$$
As $\rho_A>0$, so we have
$$\begin{align*} \sqrt{x+iy} &= e^{\tfrac12 \log(x+iy)} \\ &= e^{\tfrac12 \ln|x+iy| + i\tfrac12 \arg(x+iy)} \\ &= \sqrt[4]{x^2+y^2} \, e^{i\tfrac12 \tan^{-1}\left(\tfrac yx\right)} \\ &= \sqrt[4]{x^2+y^2} \, \left(\cos\left(\frac12 \tan^{-1}\left(\frac yx\right)\right) + i \sin \left(\frac12 \tan^{-1}\left(\frac yx\right)\right)\right) \\ &= \sqrt{\frac{x+\sqrt{x^2+y^2}}2} + i \frac{y}{\sqrt{2x+2\sqrt{x^2+y^2}}} \end{align*}$$
where $x=\sqrt2\,\rho_A$ and $y=x+\frac{x^2}2=\sqrt2\,\rho_A+\rho_A^2$ (both positive) denote the real and imaginary parts of the initial radicand. In terms of $x$, our equation is
$$\left|1 + \frac x2 + i \frac x2 + \sqrt{x + i\left(x + \frac{x^2}2\right)}\right| = R$$
and the complex square root can be expanded to
$$\left|1 + \frac x2 + i \frac x2 + \sqrt{\frac{x+\sqrt{x^2+\left(x+\frac{x^2}2\right)^2}}2} + i \frac{x+\frac{x^2}2}{\sqrt{2x+2\sqrt{x^2+\left(x+\frac{x^2}2\right)^2}}}\right| = R$$
Now take the modulus and solve for $x$... I'll leave that as an exercise for the reader, as I think this is a sufficiently convincing demonstration of just how complicated the task can get.
To the plot I've added a ray indicating the choice of $\varphi$ relative to $z_0$. The smaller circle has radius $\rho_D$ and the larger one has radius $\rho_A$.