Answer
$-1$
Work Step by Step
See p. 606,
If $z=r(\cos\theta+i\sin\theta)$, then for any integer $n$
$z^{n}=r^{n}(\cos n\theta+i\sin n\theta)$
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$z=\displaystyle \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$
$\theta$ terminates in quadrant I,
$\displaystyle \tan\theta=\frac{\sqrt{2}}{\sqrt{2}}=1$ $\Rightarrow \displaystyle \theta=\frac{\pi}{4}$,
$r=\sqrt{\frac{2}{4}+\frac{2}{4}}=\sqrt{1}=1$
$z=\displaystyle \cos\frac{\pi}{4}+i\sin\frac{\pi}{4}$
$z^{12}=1^{12}[\displaystyle \cos(12\cdot\frac{\pi}{4})+i\sin(12\cdot\frac{\pi}{4})]$
$=1(\cos 3\pi+i\sin 3\pi)$
$=-1-0$
$=-1$
