For $n∈ N$, determine the real part of $(1 + i\sqrt{3})^{n}$ For $n∈ N$, determine the real part of $(1 + i\sqrt{3})^{n}$. 
I just can't find the regularity within it. Thanks.

Use De Moivre's formula
\begin{align}
\left[ 1+i\cdot \sqrt{3} \right]^n = 
&
2^n \left[ \frac{1}{2}+i\cdot \frac{\sqrt{3}}{2} \right]^n
&
\\
=
&
2^n\cdot \left[\cos\left(\frac{\pi}{3}\right)+i\cdot \sin\left(\frac{\pi}{3}\right) \right]^n
&
\\
=
&
2^n\cdot \left[\cos\left(n\cdot\frac{\pi}{3}\right)+i\cdot \sin\left(n\cdot\frac{\pi}{3}\right) \right]
&   
\\
\end{align}
Let's $n= 3\cdot [2\cdot k]+r$. We have $r\in\{0,1,2,3,4,5\}$ and
\begin{align}
\left[ 1+i\cdot \sqrt{3} \right]^n = 
&
2^n\cdot \left[\cos\left(r\cdot\frac{\pi}{3}\right)+i\cdot \sin\left(r\cdot\frac{\pi}{3}\right) \right]
\\
\end{align}
Then we have,
$$
Re\Big(\left[ 1+i\cdot \sqrt{3} \right]^{3(2k)+r}\Big)=
\begin{cases}
2^n\cdot \left(+\frac{1}{1} \right) & \mbox{ if } r=0 \\
2^n\cdot \left(+\frac{1}{2} \right) & \mbox{ if } r=1,5 \\
2^n\cdot \left(-\frac{1}{2} \right) & \mbox{ if } r=2,4 \\
2^n\cdot \left(-\frac{1}{1} \right) & \mbox{ if } r=3 \\
\end{cases}
$$
