Let $n \in \mathbb{N}, n \ge 3.$
$a)$ Prove that there exist $z_1,z_2,…,z_n \in \mathbb{C}$ such that $$\frac{z_1}{z_2}+ \frac{z_2}{z_3}+…+ \frac{z_{n-1}}{z_n}+ \frac{z_n}{z_1}=n \mathrm{i}.$$
$b)$ Which are the values of $n$ for which there exist the complex numbers $z_1,z_2,…,z_n,$ of the same modulus, such that $$\frac{z_1}{z_2}+ \frac{z_2}{z_3}+…+ \frac{z_{n-1}}{z_n}+ \frac{z_n}{z_1}=n \mathrm{i}?$$

1. **Part (a):** Prove that there exist \( z_1, z_2, \ldots, z_n \in \mathbb{C} \) such that 
   \[
   \frac{z_1}{z_2} + \frac{z_2}{z_3} + \cdots + \frac{z_{n-1}}{z_n} + \frac{z_n}{z_1} = n \mathrm{i}.
   \]

   Let's consider the quadratic equation \( Z^2 + (n-2-ni)Z + 1 = 0 \). We need to find a root \( z \) of this equation. The roots of the quadratic equation \( AZ^2 + BZ + C = 0 \) are given by:
   \[
   Z = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}.
   \]
   For our equation, \( A = 1 \), \( B = n-2-ni \), and \( C = 1 \). Thus, the roots are:
   \[
   Z = \frac{-(n-2-ni) \pm \sqrt{(n-2-ni)^2 - 4}}{2}.
   \]
   Simplifying the discriminant:
   \[
   (n-2-ni)^2 - 4 = (n-2)^2 - 2ni(n-2) - n^2 + 4n^2i^2 - 4 = n^2 - 4n + 4 - 2ni(n-2) - n^2 + 4n^2(-1) - 4 = -4n + 4 - 2ni(n-2) - 4n^2 - 4.
   \]
   This simplifies to:
   \[
   -4n - 2ni(n-2) - 4n^2.
   \]
   Therefore, the roots are:
   \[
   Z = \frac{-(n-2-ni) \pm \sqrt{-4n - 2ni(n-2) - 4n^2}}{2}.
   \]
   We choose one of these roots, say \( z \). Now, let \( (z_1, z_2, \ldots, z_{n-1}, z_n) = (1, 1, \ldots, 1, z) \). Then:
   \[
   \frac{z_1}{z_2} + \frac{z_2}{z_3} + \cdots + \frac{z_{n-1}}{z_n} + \frac{z_n}{z_1} = 1 + 1 + \cdots + 1 + \frac{z}{1} = n-1 + z.
   \]
   We need \( n-1 + z = n \mathrm{i} \), so \( z = n \mathrm{i} - (n-1) \). This choice of \( z \) satisfies the equation.

2. **Part (b):** Determine the values of \( n \) for which there exist complex numbers \( z_1, z_2, \ldots, z_n \) of the same modulus such that 
   \[
   \frac{z_1}{z_2} + \frac{z_2}{z_3} + \cdots + \frac{z_{n-1}}{z_n} + \frac{z_n}{z_1} = n \mathrm{i}.
   \]

   Suppose \( |z_1| = |z_2| = \cdots = |z_n| = r \). Then each \( \frac{z_k}{z_{k+1}} \) has modulus 1. Let \( z_k = r e^{i \theta_k} \). Then:
   \[
   \frac{z_k}{z_{k+1}} = e^{i (\theta_k - \theta_{k+1})}.
   \]
   The sum becomes:
   \[
   e^{i (\theta_1 - \theta_2)} + e^{i (\theta_2 - \theta_3)} + \cdots + e^{i (\theta_{n-1} - \theta_n)} + e^{i (\theta_n - \theta_1)} = n \mathrm{i}.
   \]
   Each term \( e^{i (\theta_k - \theta_{k+1})} \) must be \( i \) for the sum to be \( n \mathrm{i} \). Therefore, \( \theta_k - \theta_{k+1} = \frac{\pi}{2} \) (mod \( 2\pi \)). This implies:
   \[
   \theta_1 - \theta_2 = \frac{\pi}{2}, \quad \theta_2 - \theta_3 = \frac{\pi}{2}, \quad \ldots, \quad \theta_{n-1} - \theta_n = \frac{\pi}{2}, \quad \theta_n - \theta_1 = \frac{\pi}{2}.
   \]
   Summing these equations:
   \[
   (\theta_1 - \theta_2) + (\theta_2 - \theta_3) + \cdots + (\theta_{n-1} - \theta_n) + (\theta_n - \theta_1) = n \cdot \frac{\pi}{2} = 2k\pi,
   \]
   for some integer \( k \). Therefore, \( n \cdot \frac{\pi}{2} = 2k\pi \), which implies:
   \[
   n = 4k.
   \]
   Hence, \( n \) must be a multiple of 4.

The final answer is \( \boxed{ n \equiv 0 \pmod{4} } \).