1. **State the problem:** We need to solve the integral $$\int \left(1-\frac{1}{z}\right)^2 \, dz$$. 2. **Expand the integrand:** Use the formula for the square of a binomial: $$(a-b)^2 = a^2 - 2ab + b^2$$. Here, $a=1$ and $b=\frac{1}{z}$, so: $$\left(1-\frac{1}{z}\right)^2 = 1^2 - 2 \cdot 1 \cdot \frac{1}{z} + \left(\frac{1}{z}\right)^2 = 1 - \frac{2}{z} + \frac{1}{z^2}$$ 3. **Rewrite the integral:** $$\int \left(1 - \frac{2}{z} + \frac{1}{z^2}\right) dz = \int 1 \, dz - 2 \int \frac{1}{z} \, dz + \int \frac{1}{z^2} \, dz$$ 4. **Integrate each term separately:** - $$\int 1 \, dz = z$$ - $$\int \frac{1}{z} \, dz = \ln|z|$$ - $$\int \frac{1}{z^2} \, dz = \int z^{-2} \, dz = \frac{z^{-1}}{-1} = -\frac{1}{z}$$ 5. **Combine the results:** $$z - 2 \ln|z| - \frac{1}{z} + C$$ where $C$ is the constant of integration. **Final answer:** $$\int \left(1-\frac{1}{z}\right)^2 \, dz = z - 2 \ln|z| - \frac{1}{z} + C$$