Determine all pairs of polynomials $f,g\in\mathbb{C}[x]$ with complex coefficients such that the following equalities hold for all $x\in\mathbb{C}$:

$f(f(x))-g(g(x))=1+i$
$f(g(x))-g(f(x))=1-i$
1. **Given Equations:**
   We start with the given equations for polynomials \( f \) and \( g \) with complex coefficients:
   \[
   f(f(x)) - g(g(x)) = 1 + i
   \]
   \[
   f(g(x)) - g(f(x)) = 1 - i
   \]

2. **Divisibility Argument:**
   Consider the polynomial \( f(x) - g(x) \). We know that:
   \[
   f(x) - g(x) \mid f(f(x)) - f(g(x))
   \]
   and
   \[
   f(x) - g(x) \mid g(f(x)) - g(g(x))
   \]

3. **Combining the Equations:**
   Adding and subtracting the given equations, we get:
   \[
   (f(f(x)) - f(g(x))) + (g(f(x)) - g(g(x))) = (f(f(x)) - g(g(x))) - (f(g(x)) - g(f(x)))
   \]
   Substituting the given values:
   \[
   (f(f(x)) - g(g(x))) - (f(g(x)) - g(f(x))) = (1 + i) - (1 - i) = 2i
   \]

4. **Constant Difference:**
   Since \( f(x) - g(x) \) divides \( 2i \), it must be a constant polynomial. Let \( f(x) - g(x) = k \), where \( k \) is a constant. Plugging this back into the given relations, we find:
   \[
   f(x) = g(x) + k
   \]

5. **Determining \( k \):**
   Substitute \( f(x) = g(x) + k \) into the first equation:
   \[
   f(f(x)) = f(g(x) + k) = g(g(x) + k) + k
   \]
   \[
   g(g(x) + k) + k - g(g(x)) = 1 + i
   \]
   Since \( g(g(x) + k) - g(g(x)) \) must be a constant, the degree of \( g \) must be 1. Let \( g(x) = ax + b \).

6. **Form of \( g(x) \):**
   Since \( g \) is a linear polynomial, we write:
   \[
   g(x) = ax + b
   \]

7. **Form of \( f(x) \):**
   Since \( f(x) = g(x) + k \), we have:
   \[
   f(x) = ax + b + k
   \]

8. **Substituting into the Equations:**
   Substitute \( f(x) = ax + b + k \) and \( g(x) = ax + b \) into the second equation:
   \[
   f(g(x)) = f(ax + b) = a(ax + b) + b + k = a^2x + ab + b + k
   \]
   \[
   g(f(x)) = g(ax + b + k) = a(ax + b + k) + b = a^2x + ab + ak + b
   \]
   \[
   f(g(x)) - g(f(x)) = (a^2x + ab + b + k) - (a^2x + ab + ak + b) = k - ak = 1 - i
   \]
   Solving for \( k \):
   \[
   k(1 - a) = 1 - i
   \]
   If \( k = 1 \), then:
   \[
   1(1 - a) = 1 - i \implies 1 - a = 1 - i \implies a = i
   \]

9. **Final Polynomials:**
   With \( a = i \) and \( k = 1 \), we have:
   \[
   g(x) = ix + b
   \]
   \[
   f(x) = ix + b + 1
   \]
   Substitute into the first equation to find \( b \):
   \[
   f(f(x)) = f(ix + b + 1) = i(ix + b + 1) + b + 1 = -x + ib + i + b + 1
   \]
   \[
   g(g(x)) = g(ix + b) = i(ix + b) + b = -x + ib + b
   \]
   \[
   f(f(x)) - g(g(x)) = (-x + ib + i + b + 1) - (-x + ib + b) = i + 1
   \]
   This holds for any \( b \).

The final answer is \( \boxed{ f(x) = ix + 1 } \) and \( g(x) = ix \).