1. The problem asks to determine the values of $\ln(-1 - i)$ and its principal value. 2. Recall that for a complex number $z = re^{i\theta}$, the complex logarithm is given by: $$\ln(z) = \ln(r) + i\theta$$ where $r = |z|$ is the modulus and $\theta = \arg(z)$ is the argument (angle) of $z$. 3. First, find the modulus $r$ of $z = -1 - i$: $$r = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$$ 4. Next, find the argument $\theta$ of $z$. Since $z$ is in the third quadrant (both real and imaginary parts negative), $$\theta = \pi + \arctan\left(\frac{-1}{-1}\right) = \pi + \arctan(1) = \pi + \frac{\pi}{4} = \frac{5\pi}{4}$$ 5. Therefore, the complex logarithm is: $$\ln(-1 - i) = \ln\left(\sqrt{2}\right) + i \frac{5\pi}{4} = \frac{1}{2} \ln(2) + i \frac{5\pi}{4}$$ 6. The principal value of the logarithm is defined with the argument $\theta$ in $(-\pi, \pi]$. Since $\frac{5\pi}{4} > \pi$, subtract $2\pi$ to bring it into the principal range: $$\theta_{principal} = \frac{5\pi}{4} - 2\pi = -\frac{3\pi}{4}$$ 7. Hence, the principal value of $\ln(-1 - i)$ is: $$\ln(-1 - i) = \frac{1}{2} \ln(2) - i \frac{3\pi}{4}$$ Final answers: - All values: $$\ln(-1 - i) = \frac{1}{2} \ln(2) + i \frac{5\pi}{4} + 2\pi i k, \quad k \in \mathbb{Z}$$ - Principal value: $$\ln(-1 - i) = \frac{1}{2} \ln(2) - i \frac{3\pi}{4}$$