1. The problem is to find the value of $$z = \frac{\log(1+i)}{1-i}$$ where $i$ is the imaginary unit with $i^2 = -1$. 2. Recall that $\log$ of a complex number $z = re^{i\theta}$ (in polar form) is given by $$\log z = \log r + i\theta$$ where $r = |z|$ and $\theta = \arg(z)$. 3. First, find $1+i$ in polar form: - Magnitude: $$r = |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2}$$ - Argument: $$\theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}$$ 4. So, $$\log(1+i) = \log\sqrt{2} + i\frac{\pi}{4} = \frac{1}{2}\log 2 + i\frac{\pi}{4}$$ 5. Next, simplify the denominator $1 - i$ by multiplying numerator and denominator by the conjugate $1 + i$: $$\frac{\log(1+i)}{1-i} \times \frac{1+i}{1+i} = \frac{\log(1+i)(1+i)}{(1-i)(1+i)}$$ 6. Calculate the denominator: $$(1-i)(1+i) = 1 - i^2 = 1 - (-1) = 2$$ 7. Calculate the numerator: $$\log(1+i)(1+i) = \left(\frac{1}{2}\log 2 + i\frac{\pi}{4}\right)(1+i)$$ 8. Multiply out: $$= \frac{1}{2}\log 2 \times 1 + \frac{1}{2}\log 2 \times i + i\frac{\pi}{4} \times 1 + i\frac{\pi}{4} \times i$$ $$= \frac{1}{2}\log 2 + i\frac{1}{2}\log 2 + i\frac{\pi}{4} + i^2 \frac{\pi}{4}$$ 9. Since $i^2 = -1$, this becomes: $$= \frac{1}{2}\log 2 + i\frac{1}{2}\log 2 + i\frac{\pi}{4} - \frac{\pi}{4}$$ 10. Group real and imaginary parts: $$= \left(\frac{1}{2}\log 2 - \frac{\pi}{4}\right) + i\left(\frac{1}{2}\log 2 + \frac{\pi}{4}\right)$$ 11. Divide by denominator 2: $$z = \frac{1}{2} \left(\frac{1}{2}\log 2 - \frac{\pi}{4}\right) + i \frac{1}{2} \left(\frac{1}{2}\log 2 + \frac{\pi}{4}\right)$$ 12. Simplify: $$z = \frac{1}{4}\log 2 - \frac{\pi}{8} + i \left( \frac{1}{4}\log 2 + \frac{\pi}{8} \right)$$ Final answer: $$\boxed{z = \frac{1}{4}\log 2 - \frac{\pi}{8} + i \left( \frac{1}{4}\log 2 + \frac{\pi}{8} \right)}$$