Let $i$ be a root of the equation $x^2+1=0$ and let $\omega$ be a root of the equation $x^2+x+1=0$ . Construct a polynomial $$f(x)=a_0+a_1x+\cdots+a_nx^n$$ where $a_0,a_1,\cdots,a_n$ are all integers such that $f(i+\omega)=0$ .
1. First, identify the roots of the given equations:
   - The equation \(x^2 + 1 = 0\) has roots \(i\) and \(-i\).
   - The equation \(x^2 + x + 1 = 0\) has roots \(\omega\) and \(\omega^2\), where \(\omega = e^{2\pi i / 3}\) and \(\omega^2 = e^{-2\pi i / 3}\).

2. We need to construct a polynomial \(f(x)\) with integer coefficients such that \(f(i + \omega) = 0\).

3. Consider the conjugates of \(i\) and \(\omega\):
   - The conjugate of \(i\) is \(-i\).
   - The conjugates of \(\omega\) are \(\omega\) and \(\omega^2\).

4. The sum \(i + \omega\) is a complex number. To ensure the polynomial has integer coefficients, we need to consider the conjugate pairs:
   - The conjugate of \(i + \omega\) is \(-i + \omega^2\).

5. Construct the polynomial whose roots are \(i + \omega\) and \(-i + \omega^2\):
   \[
   (x - (i + \omega))(x - (-i + \omega^2))
   \]

6. Expand the product:
   \[
   (x - (i + \omega))(x - (-i + \omega^2)) = (x - i - \omega)(x + i - \omega^2)
   \]

7. Use the fact that \(\omega^2 = \overline{\omega}\) and \(\omega + \omega^2 = -1\):
   \[
   (x - i - \omega)(x + i - \omega^2) = (x - i - \omega)(x + i - \overline{\omega})
   \]

8. Expand the product:
   \[
   (x - i - \omega)(x + i - \overline{\omega}) = x^2 + x(\overline{\omega} - \omega) + i(\overline{\omega} - \omega) - i^2 - \omega \overline{\omega}
   \]

9. Simplify the expression:
   \[
   x^2 + x(\overline{\omega} - \omega) + i(\overline{\omega} - \omega) - (-1) - 1 = x^2 + x(\overline{\omega} - \omega) + i(\overline{\omega} - \omega) + 2
   \]

10. Since \(\overline{\omega} - \omega = -\sqrt{3}i\) and \(i(\overline{\omega} - \omega) = -\sqrt{3}\):
    \[
    x^2 + x(-\sqrt{3}i) - \sqrt{3} + 2
    \]

11. To ensure integer coefficients, we need to square the polynomial:
    \[
    (x^2 + x + 2)^2 = x^4 + 2x^3 + 5x^2 + 4x + 4
    \]

12. Therefore, the polynomial \(f(x)\) with integer coefficients such that \(f(i + \omega) = 0\) is:
    \[
    f(x) = x^4 + 2x^3 + 5x^2 + 4x + 4
    \]

The final answer is \(\boxed{x^4 + 2x^3 + 5x^2 + 4x + 4}\)