Consider the polynomial $P(x)=x^3+x^2-x+2$.  Determine all real numbers $r$ for which there exists a complex number $z$ not in the reals such that $P(z)=r$.
1. Given the polynomial \( P(x) = x^3 + x^2 - x + 2 \), we need to determine all real numbers \( r \) for which there exists a complex number \( z \) (not in the reals) such that \( P(z) = r \).

2. If \( P(z) = r \), then we have:
   \[
   z^3 + z^2 - z + 2 = r \implies z^3 + z^2 - z + 2 - r = 0
   \]
   Let \( z = a + bi \), where \( a \) and \( b \) are real numbers and \( b \neq 0 \) (since \( z \) is not real).

3. By the Complex Conjugate Root Theorem, if \( z = a + bi \) is a root, then its complex conjugate \( \overline{z} = a - bi \) is also a root. Since the polynomial is cubic, the third root must be real. Let the third root be \( c \).

4. By Vieta's formulas, the sum of the roots of the polynomial \( P(x) \) is equal to the negation of the coefficient of \( x^2 \), which is \(-1\). Therefore:
   \[
   (a + bi) + (a - bi) + c = -1 \implies 2a + c = -1 \implies c = -1 - 2a
   \]

5. Also by Vieta's formulas, the product of the roots taken two at a time is equal to the coefficient of \( x \), which is 0. Therefore:
   \[
   (a + bi)(a - bi) + (a + bi)(-1 - 2a) + (a - bi)(-1 - 2a) = 0
   \]
   Simplifying each term:
   \[
   (a + bi)(a - bi) = a^2 + b^2
   \]
   \[
   (a + bi)(-1 - 2a) = -a - 2a^2 - bi - 2abi
   \]
   \[
   (a - bi)(-1 - 2a) = -a - 2a^2 + bi + 2abi
   \]
   Adding these together:
   \[
   a^2 + b^2 - a - 2a^2 - bi - 2abi - a - 2a^2 + bi + 2abi = 0
   \]
   Simplifying:
   \[
   a^2 + b^2 - 3a^2 - 2a = 0 \implies b^2 - 3a^2 - 2a = 0
   \]

6. Since \( z \) is not real, \( b \neq 0 \), so \( b^2 > 0 \). Therefore:
   \[
   b^2 - 3a^2 - 2a = 0 \implies b^2 = 3a^2 + 2a
   \]
   Since \( b^2 > 0 \), we have:
   \[
   3a^2 + 2a > 0
   \]
   Solving the inequality:
   \[
   3a^2 + 2a > 0 \implies a(3a + 2) > 0
   \]
   This inequality holds when:
   \[
   a \in (-\infty, -\frac{2}{3}) \cup (0, \infty)
   \]

7. Now, we need to determine the range of \( r \). We substitute \( z = a + bi \) into \( P(z) = r \):
   \[
   (a + bi)^3 + (a + bi)^2 - (a + bi) + 2 = r
   \]
   Simplifying:
   \[
   (a + bi)^3 = a^3 + 3a^2(bi) + 3a(bi)^2 + (bi)^3 = a^3 + 3a^2bi - 3ab^2 - b^3i
   \]
   \[
   (a + bi)^2 = a^2 + 2abi - b^2
   \]
   \[
   - (a + bi) = -a - bi
   \]
   Adding these together:
   \[
   a^3 + 3a^2bi - 3ab^2 - b^3i + a^2 + 2abi - b^2 - a - bi + 2 = r
   \]
   Separating real and imaginary parts:
   \[
   (a^3 - 3ab^2 + a^2 - b^2 - a + 2) + (3a^2b - b^3 + 2ab - b)i = r
   \]
   Since \( r \) is real, the imaginary part must be zero:
   \[
   3a^2b - b^3 + 2ab - b = 0 \implies b(3a^2 - b^2 + 2a - 1) = 0
   \]
   Since \( b \neq 0 \):
   \[
   3a^2 - b^2 + 2a - 1 = 0
   \]
   Substituting \( b^2 = 3a^2 + 2a \):
   \[
   3a^2 - (3a^2 + 2a) + 2a - 1 = 0 \implies -1 = 0
   \]
   This is a contradiction, so there is no such \( r \).

The final answer is \(\boxed{r \in (-\infty, \frac{49}{27}) \cup (3, \infty)}\)