2
$\begingroup$

Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ,$ find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}.$

Solution: We have $z=e^{i\theta}$, so $e^{i\theta}+\frac{1}{e^{i\theta}}=\frac{\cos \theta}{\cos ^2\theta +\sin ^2\theta}+\cos \theta +i\sin \theta - \frac{i\sin \theta}{\cos ^2\theta +\sin ^2\theta}=2\cos \theta$. Therefore, $\theta =3^\circ=\frac{\pi}{60}$. From there, $z^{2000}+\frac{1}{z^{2000}}=e^{\frac{100\pi (i)}{3}}+e^{\frac{-100\pi (i)}{3}}=2\cos \frac{4\pi}{3}=-1$, so our answer is $\boxed{0}$.

How to solve this by applying Tchebyshev?

$\endgroup$
2
  • $\begingroup$ @AnuragA Euler's Identity $\endgroup$
    – Sullo
    Dec 4, 2019 at 21:34
  • $\begingroup$ Do you mean one of these theorems, or something else? $\endgroup$
    – J.G.
    Dec 4, 2019 at 21:59

1 Answer 1

3
$\begingroup$

Note that, given $z+\frac1z = 2\cos a$ for any $a$,

$$z^n+\frac1{z^n}=2\cos (na)$$

holds true, which is shown recursively below,

$$z^n+\frac1{z^n}= \left(z+\frac1z\right)\left(z^{n-1}+\frac1{z^{n-1}}\right) -\left(z^{n-2}+\frac1{z^{n-2}}\right)$$ $$=2\cos a \cdot 2\cos [(n-1)a] - 2\cos [(n-2)a] $$ $$=2[\cos (na) + \cos [(n-2)a]]- 2\cos [(n-2)a] =2\cos (na) $$

Thus, for $a=3^\circ=\frac\pi{60}$,

$$z^{2000}+\frac 1{z^{2000}}=2\cos\left(\frac{2000\pi}{60}\right)=-1$$

$\endgroup$

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.