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

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

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I am a little confused by your formula. Is it "one plus i times square root of 3" in the parentheses? – Code-Guru Jan 8 '13 at 21:43
@Code-Guru - this is how I read it too – Belgi Jan 8 '13 at 21:43
Regularities will come quickly if you compute. Maybe powers of $\frac{1}{2}+\frac{\sqrt{3}}{2}i$ will be more familiar. – André Nicolas Jan 8 '13 at 22:26

2 Answers 2

Hint: Consider polar representation

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

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For questions tagged (homework), try to avoid giving complete answers. See this for more informtation. – robjohn Jan 8 '13 at 22:26

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