Problem 1333

1. Interpret the inequalities geometrically:
- The inequality $|z - 1| < 1$ describes the interior of a unit circle centered at $1$ in the complex plane.
- The inequality $\left| \frac{1}{z} - 1 \right| < 1$ can be rewritten by clearing the denominator. Let $z = x + yi$ where $x$ and $y$ are real numbers.

2. Rewrite the first inequality:
$$\left| \frac{1}{z} - 1 \right| < 1 \implies \left| \frac{1 - z}{z} \right| < 1 \implies |1 - z| < |z|$$
This inequality $|1 - z| < |z|$ is satisfied by all $z$ with real part $\Re(z) > \frac{1}{2}$. This can be shown by considering the geometric interpretation of the magnitudes.

3. Combine the regions:
- The region $|z - 1| < 1$ is a unit circle centered at $1$.
- The region $|1 - z| < |z|$ is the half-plane where the real part of $z$ is greater than $\frac{1}{2}$.

4. Find the intersection:
- The intersection of these two regions is the part of the unit circle centered at $1$ that lies to the right of the vertical line $\Re(z) = \frac{1}{2}$.

5. Calculate the area:
- The intersection forms a sector of the circle and a triangle.
- The angle of the sector is $\frac{2\pi}{3}$ radians (since the line $\Re(z) = \frac{1}{2}$ intersects the circle at angles $\pm \frac{\pi}{3}$ from the positive real axis).
- The area of the sector is:
$$\text{Area of sector} = \frac{1}{2} \times (\text{radius})^2 \times (\text{angle}) = \frac{1}{2} \times 1^2 \times \frac{2\pi}{3} = \frac{\pi}{3}$$
- The area of the triangle formed by the intersection points and the center of the circle is:
$$\text{Area of triangle} = \frac{1}{2} \times (\text{base}) \times (\text{height}) = \frac{1}{2} \times 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}$$

6. Sum the areas:
$$\text{Total area} = \frac{\pi}{3} + \frac{\sqrt{3}}{4}$$

The final answer is $\boxed{\frac{\pi}{3} + \frac{\sqrt{3}}{4}}$