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here's a question I haven't been able to solve.

Let $z_{n}=(-\frac{3}{2}+\frac{\sqrt{3}}{2}i)^{n}$. Find the least positive integer $n$ such that $|z_{n+1}-z_{n}|^{2}>7000$.

Ok so far I've done some really messy working using the addition formulas for $|(\sqrt{3})^{n+1}(\cos\frac{5(n+1)\pi}{6}+i\sin\frac{5(n+1)\pi}{6})-(\sqrt{3})^{n}(\cos\frac{5n\pi}{6}+i\sin\frac{5n\pi}{6})|^{2}>7000$ and I've gotten $|(\sqrt{3})^{n}(-\frac{5}{2}\cos\frac{5n\pi}{6}-\frac{\sqrt{3}}{2}\sin\frac{5n\pi}{6}-i\frac{5}{2}\sin\frac{5n\pi}{6}+i\frac{\sqrt{3}}{2}\cos\frac{5n\pi}{6})|^{2}>7000$, which looks kind of simpler I guess, if I didn't make any mistakes. But I'm not very sure how to continue or if this is not the right method because I'm just using brute force and I don't really see any trick anywhere. Could someone provide a hint? Thanks!

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Let $\alpha = -\frac{3}{2} + \frac{\sqrt{3}}{2}i$. We have $z_{n+1}-z_n = \alpha^n (\alpha-1)$, and so $|z_{n+1}-z_n| = |\alpha^n (\alpha-1)| = |\alpha|^n \cdot|\alpha - 1|$, since multiplying complex numbers multiplies their modulus. You now solve $|\alpha|^n \cdot|\alpha - 1| > \sqrt{7000}$ which after computing $|\alpha|$ and $|\alpha-1|$, and then take the smallest integer solution, which happens to be $n = \lfloor \log_{|\alpha|} \left( \frac{\sqrt{7000}}{|\alpha - 1|} \right) \rfloor + 1$.

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  • $\begingroup$ sorry there is an edit because i missed out a square but i think the method should still be similar, thanks $\endgroup$ – A. Lim May 21 at 15:34
  • $\begingroup$ @auscrypt Good answer and +1. It woos be better if you substitute $|\alpha|=3$ in your answer. $\endgroup$ – Qurultay May 21 at 15:37
  • $\begingroup$ @A.Lim No problem, I've edited it just in case. $\endgroup$ – auscrypt May 21 at 15:37

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