For $n∈ N$, determine the real part of $(1 + i\sqrt{3})^{n}$

Question

For $n∈ N$, determine the real part of $(1 + i\sqrt{3})^{n}$.

I just can't find the regularity within it. Thanks.

Answer 1

Hint: Consider polar representation

Answer 2

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, $$ Re\Big(\left[ 1+i\cdot \sqrt{3} \right]^{3(2k)+r}\Big)= \begin{cases} 2^n\cdot \left(+\frac{1}{1} \right) & \mbox{ if } r=0 \\ 2^n\cdot \left(+\frac{1}{2} \right) & \mbox{ if } r=1,5 \\ 2^n\cdot \left(-\frac{1}{2} \right) & \mbox{ if } r=2,4 \\ 2^n\cdot \left(-\frac{1}{1} \right) & \mbox{ if } r=3 \\ \end{cases} $$