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Number theory Difficulty 4.9 AIME Find the answer

2. Let the 20 vertices of a regular 20-sided polygon inscribed in the unit circle in the complex plane correspond to the complex numbers z1,z2,⋯ ,z20z_{1}, z_{2}, \cdots, z_{20}. Then the number of different points corresponding to the complex numbers z11995z_{1}^{1995}, z21905,⋯ ,z201195z_{2}^{1905}, \cdots, z_{20}^{1195} is:

Pick one

Solution

2. (A).

Let z1=cos⁡θ+isin⁡θz_{1}=\cos \theta+i \sin \theta, then
zt=(cos⁡θ+isin⁡θ)(cos⁡2(k−1)π20+isin⁡2(k−1)π20),1⩽k⩽20. \begin{aligned} z_{t}= & (\cos \theta+i \sin \theta)\left(\cos \frac{2(k-1) \pi}{20}\right. \\ & \left.+i \sin \frac{2(k-1) \pi}{20}\right), 1 \leqslant k \leqslant 20 . \end{aligned}

From 1995=20×99+151995=20 \times 99+15, we get
zk995=(cos⁡1995θ+isin⁡1995θ)(cos⁡3π2+isin⁡3π2)k−1=(cos⁡1995θ+isin⁡1995θ)(−i)k−1, \begin{aligned} z k^{995}= & (\cos 1995 \theta+i \sin 1995 \theta) \\ & \left(\cos \frac{3 \pi}{2}+i \sin \frac{3 \pi}{2}\right)^{k-1} \\ = & (\cos 1995 \theta+i \sin 1995 \theta)(-i)^{k-1}, \end{aligned}
k=1,2,⋯ ,20k=1,2, \cdots, 20. There are four different values.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.