Given the equation $x^2+ax+1=0$, determine:

a) The interval of possible values for $a$ where the solutions to the previous equation are not real.
b) The loci of the roots of the polynomial, when $a$ is in the previous interval.
1. **Determine the interval of possible values for \( a \) where the solutions to the equation \( x^2 + ax + 1 = 0 \) are not real.**

   To determine when the solutions are not real, we need to analyze the discriminant of the quadratic equation. The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by:
   \[
   \Delta = b^2 - 4ac
   \]
   For the given equation \(x^2 + ax + 1 = 0\), we have \(a = 1\), \(b = a\), and \(c = 1\). Thus, the discriminant is:
   \[
   \Delta = a^2 - 4 \cdot 1 \cdot 1 = a^2 - 4
   \]
   The solutions are not real if the discriminant is less than zero:
   \[
   a^2 - 4 < 0
   \]
   Solving this inequality:
   \[
   a^2 < 4
   \]
   Taking the square root of both sides:
   \[
   -2 < a < 2
   \]
   Therefore, the interval of possible values for \( a \) where the solutions are not real is:
   \[
   a \in (-2, 2)
   \]

2. **Determine the loci of the roots of the polynomial when \( a \) is in the interval \((-2, 2)\).**

   The roots of the quadratic equation \(x^2 + ax + 1 = 0\) can be found using the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   For our equation \(x^2 + ax + 1 = 0\), the roots are:
   \[
   x = \frac{-a \pm \sqrt{a^2 - 4}}{2}
   \]
   When \(a \in (-2, 2)\), the discriminant \(a^2 - 4\) is negative, so the roots are complex. Let \(a = 2i\sin(\theta)\) where \(\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). Then the roots are:
   \[
   x = \frac{-2i\sin(\theta) \pm \sqrt{(2i\sin(\theta))^2 - 4}}{2}
   \]
   Simplifying inside the square root:
   \[
   x = \frac{-2i\sin(\theta) \pm \sqrt{-4\sin^2(\theta) - 4}}{2}
   \]
   \[
   x = \frac{-2i\sin(\theta) \pm 2i\sqrt{\sin^2(\theta) + 1}}{2}
   \]
   \[
   x = -i\sin(\theta) \pm i\sqrt{\sin^2(\theta) + 1}
   \]
   Since \(\sin^2(\theta) + 1 = \cos^2(\theta) + \sin^2(\theta) = 1\), we have:
   \[
   x = -i\sin(\theta) \pm i
   \]
   Let \(x = u + iv\), then:
   \[
   u = 0, \quad v = -\sin(\theta) \pm 1
   \]
   The loci of the roots are points on the complex plane where the imaginary part \(v\) satisfies:
   \[
   v = -\sin(\theta) \pm 1
   \]
   This describes an ellipse in the complex plane. The equation of the ellipse is:
   \[
   4x^2 + y^2 = 4
   \]
   except at the points \((-1, 0)\) and \((1, 0)\).

The final answer is \( \boxed{ a \in (-2, 2) } \) and the loci of the roots is the ellipse \(4x^2 + y^2 = 4\) except points \((-1, 0)\) and \((1, 0)\).