Wolfram Alpha disagrees with my computation and my first guess is that is because a different branch cut is chosen, but this doesn't seem to be the case after checking, so I'm curious.
I have
\begin{align}
(\sqrt{2} + \sqrt{2}i)^{1+i} &= \exp\big( (1+i) \ln \big( \sqrt{2}(1+i) \big) \big) \\
&= \exp \big( (1+i)\ln \big(2 e^{\frac{\pi i}{4}} \big) \big) \\
&= 2\exp \big( (1+i)\frac{\pi i}{4} \big) \\
&= \frac{2}{e^{\frac{\pi}{4}}}\exp \big( \frac{\pi i}{4} \big) .
\end{align}
Wolfram Alpha however seems to compute an angle of $\theta = 84.7 ^\circ$, as opposed to $\theta = \frac{\pi}{4} = 45 ^\circ$.
The only freedom that we could have would be to pick a different branch $k$: $\ln(e^{\frac{\pi}{4} +2\pi k})$. But this shifts the the angle by $360 ^\circ$, so I'm really puzzled.
How would one reproduce Wolfram Alpha's answer? And more importantly, why do the results disagree? The choice of branch cut doesn't seem to make difference.
Answer
Let's take $\ln 2+\pi i/4$ as the principal logarithm of $\sqrt2(1+i)$.
Then
$$(1+i)\left(\ln 2+\frac{\pi i}4\right)
=\ln2-\frac\pi 4+i\left(\ln2+\frac {\pi}4\right).$$
The "principal value" you seek is the exponential of this.
Note that $\ln2+\pi/4$ radians is about $84.7$ degrees.
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