Could someone please check my result for this complex logarithm exercise? I solved the following exercise and would appreciate it if someone would check my answer:

Exercise: Find the principal value of the logarithm of the following numbers:

$(i(i-1))^i$ and $i^i (i-1)^i$

My answer:

$$ \log((i(i-1))^i) = i \log (i (i-1)) = i(\log \sqrt{2} + i \arg e^{i \pi  5 \over 4}) = i \log \sqrt{2} + {3 \over 4 }\pi$$

and

$$ \log i^i(i-1)^i = \log (e^{i \pi \over 2})^i(e^{-i \pi \over 4})^i = \log e^{-{\pi \over 2}} e^{\pi \over 4} = {\pi \over 4 } - {\pi \over 2} = - {\pi \over 4}$$
$i(i-1)=-1-i$ so both expressions are equal to $(-1-i)^i = (\sqrt 2 e^{i\frac{5\pi}{4}})^i$

$log (-1-i)^i = i \;log(\sqrt 2 e^{i\frac{5\pi}{4}})= -\frac{5 \pi}{4}+i\;log\sqrt2$  is correct for both expressions.
$i(i-1)=-1-i$ so both expressions are equal to $(-1-i)^i = (\sqrt 2 e^{i\frac{5\pi}{4}})^i$

$log (-1-i)^i = i \;log(\sqrt 2 e^{i\frac{5\pi}{4}})= -\frac{5 \pi}{4}+i\;log\sqrt2$  is correct for both expressions.