Let \(z = \sqrt{2} \operatorname{cis} 3\).
What is the real part of \(z^6\) rounded to 4 decimal places?
The answer is \(5.2825\).
Using de Moivre's formulat, we see that
\(z^6 = \sqrt{2}^6 \operatorname{cis} (6\cdot 3) = 8 \operatorname{cis} 18\).
Hence, the real part is \(8 \operatorname{cos} 18 =
5.28253366595\ldots\approx 5.2825\).