Plotting a complex argument arc I am having trouble sketching a complex argument arc

$$ \text{Sketch the following on an arcand diagram:}\\ \arg\left(\frac{w+1}{w}\right)=\frac{\pi}{6}$$

I've tried to devise a method on my own looking at questions and answers but it has failed me on this specific question so I require some help.
What I have been doing so far is I was writing $$\arg\left(\frac{x+iy+1}{x+iy}\right)$$
$$x+1=0\\y=0\\y=0 \\ x=0$$
Getting two points (-1,0) and (0,0). So I take the two points and connect them with an arc.
However the solution of the exercise was this:

However my answer was exactly the opposite on the y-axis. Can anyone show me a method and the logic behind the drawing of the arc, and the arc of any other similar question?

Taking $w = x+iy$, we get that $$\text{Arg} \left( \dfrac{1+w}w\right) = \text{Arg} \left( 1+\dfrac1w\right) = \text{Arg} \left( 1+\dfrac{x-iy}{x^2+y^2}\right) = \text{Arg} \left( \dfrac{x^2+y^2+x}{x^2+y^2} - i \dfrac{y}{x^2+y^2}\right)$$
Hence, we need $$\tan(\pi/6) = -\dfrac{y}{x^2+y^2+x}$$
$$x^2 + y^2 + x + y\sqrt{3} = 0 \text{ i.e. } \left(x + \dfrac12 \right)^2 + \left(y + \dfrac{\sqrt{3}}2 \right)^2 = 1$$
However, note that we have the argument in the first quadrant and hence $$\dfrac{x^2+y^2+x}{x^2+y^2} >0 \text{ & } -\dfrac{y}{x^2+y^2} > 0$$
This gives us that $y < 0$.
Hence, the curve we are interested in is $$\left(x + \dfrac12 \right)^2 + \left(y + \dfrac{\sqrt{3}}2 \right)^2 = 1 \text{ with } y< 0$$
