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!


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 $n = \lfloor \log_{|\alpha|} \left( \frac{\sqrt{7000}}{|\alpha - 1|} \right) \rfloor + 1$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.