Find all triplets $ (a,b,c) $ of nonzero complex numbers having the same absolute value and which verify the equality:
$$ \frac{a}{b} +\frac{b}{c}+\frac{c}{a} =-1 $$
1. Let \( x = \frac{a}{b} \), \( y = \frac{b}{c} \), and \( z = \frac{c}{a} \). Given that \( a, b, c \) are nonzero complex numbers with the same absolute value, we have \( |a| = |b| = |c| \). This implies \( |x| = |y| = |z| = 1 \).

2. From the problem, we know:
   \[
   x + y + z = -1
   \]
   Additionally, since \( x = \frac{a}{b} \), \( y = \frac{b}{c} \), and \( z = \frac{c}{a} \), we have:
   \[
   xyz = \left( \frac{a}{b} \right) \left( \frac{b}{c} \right) \left( \frac{c}{a} \right) = 1
   \]

3. Since \( |x| = |y| = |z| = 1 \), the complex conjugate of \( x \) is \( \overline{x} = \frac{1}{x} \). Similarly, \( \overline{y} = \frac{1}{y} \) and \( \overline{z} = \frac{1}{z} \). Therefore:
   \[
   \overline{x} + \overline{y} + \overline{z} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}
   \]

4. Since \( x + y + z = -1 \), taking the complex conjugate gives:
   \[
   \overline{x} + \overline{y} + \overline{z} = -1
   \]
   But we also know:
   \[
   \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = -1
   \]

5. Using the identity for the sum of reciprocals of products, we have:
   \[
   \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{xy + yz + zx}{xyz} = xy + yz + zx
   \]
   Since \( xyz = 1 \), we get:
   \[
   xy + yz + zx = -1
   \]

6. Now, \( x, y, z \) are the roots of the polynomial:
   \[
   t^3 + t^2 - t - 1 = 0
   \]
   Factoring this polynomial, we get:
   \[
   t^3 + t^2 - t - 1 = (t + 1)^2 (t - 1)
   \]
   Therefore, the roots are \( t = -1, -1, 1 \).

7. Thus, \( x, y, z \) can be \( (1, -1, -1) \) or any permutation of these values. This implies:
   \[
   \left( \frac{a}{b}, \frac{b}{c}, \frac{c}{a} \right) = (1, -1, -1)
   \]
   or any permutation.

8. Solving for \( a, b, c \), we get:
   \[
   a = b, \quad b = -c, \quad c = -a
   \]
   up to permutations. This means \( a = b \) and \( c = -a \), or any permutation of these values.

The final answer is \( \boxed{ (a, b, c) = (a, a, -a) } \) up to permutations.