/p>

the point depends on the angle $\theta$, but angles differing by a multiple of $2\pi$ give the same point.

So we use the idea:

$$ \cos \theta + i\sin \theta = \cos \phi + i\sin \phi $$

only when

$$ \theta-\phi = 2k\pi,\quad k\in \mathbb{Z} $$

Now consider set $A$:

$$ A=\left\{\cos\left(\frac{2n\pi}{5}\right)+i\sin\left(\frac{2n\pi}{5}\right):n\in\mathbb{Z}\right\} $$

Here the angle is

$$ \frac{2n\pi}{5} $$

If $n$ increases by $5$, then

$$ \frac{2(n+5)\pi}{5} = \frac{2n\pi}{5}+2\pi $$

So the point repeats after every $5$ values of $n$.

Thus $A$ has only $5$ distinct points. Hence $A$ is finite.

Now consider set $B$:

$$ B=\left\{\cos\left(\frac{2n}{5}\right)+i\sin\left(\frac{2n}{5}\right):n\in\mathbb{Z}\right\} $$

Here the angle is

$$ \frac{2n}{5} $$

For repetition, we would need

$$ \frac{2n}{5}-\frac{2m}{5}=2k\pi $$

That gives

$$ \frac{2(n-m)}{5}=2k\pi $$

Dividing by $2$,

$$ \frac{n-m}{5}=k\pi $$

So,

$$ n-m=5k\pi $$

But $n-m$ is an integer, while $5k\pi$ is not an integer unless $k=0$, because $\pi$ is irrational.

So no repetition occurs for different integer values of $n$.

Hence $B$ has infinitely many distinct points.

So $A$ is finite but $B$ is infinite.

The correct option is:

$$ \boxed{\text{Option D}} $$

Complex Numbers

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