I’m relatively new to complex numbers, and I don’t know how to solve this question. I know that $|w-3| = 2$ is a circle of radius $2$ with center $(3,0)$ but I’m not sure about the other one, and I don’t know how to find $\alpha$.
There is a unique complex number $w$ that satisfies both $|w - 3| = 2$ and $\arg(w + 1) = \alpha$, where $\alpha$ is a constant such that $0 < \alpha < \pi$.
The circle is the locus of points $w$ such that $|w-3|=2$. The upper of the two parallel lines is the locus of points $w$ such that arg($w+1)=\alpha$. The points on the lower parallel line are the points with arg $\alpha$ ($\alpha$ is the angle between the line and the real axis). Or more accurately, the points on the ray from the origin into the upper left quadrant have arg $\alpha$. The points on the other half of the line have arg outside the range $0$ to $\pi$.
You can see that if we reduce $\alpha$ from the angle shown, then the upper line intersects the circle in two points. If we increase $\alpha$ then the line does not intersect the circle.
You want the value of $\alpha$ which gives a unique $w$, so it just remains to calculate the angle. It is in a right-angled triangle. The hypotenuse is length 4 (from -1 to 3) and the opposite side is length 2 (the radius). So $\sin\alpha=\frac{1}{2}$ and hence $\alpha=30^\circ=\pi/6$.