Let be a real number $ a\in \left[ 2+\sqrt 2,4 \right] . $ Find $ \inf_{\stackrel{z\in\mathbb{C}}{|z|\le 1}} \left| z^2-az+a \right| . $
1. **Define the function and the domain:**
   We need to find the infimum of the function \( f(z) = |z^2 - az + a| \) where \( z \in \mathbb{C} \) and \( |z| \leq 1 \). 

2. **Consider the polynomial \( P(z) = z^2 - az + a \):**
   We need to analyze the behavior of \( P(z) \) on the boundary of the unit disk \( |z| = 1 \).

3. **Express \( z \) on the unit circle:**
   Let \( z = e^{i\theta} \) where \( \theta \in [0, 2\pi) \). Then,
   \[
   P(e^{i\theta}) = e^{2i\theta} - ae^{i\theta} + a.
   \]

4. **Calculate the magnitude \( |P(e^{i\theta})| \):**
   \[
   |P(e^{i\theta})| = |e^{2i\theta} - ae^{i\theta} + a|.
   \]
   Since \( |e^{i\theta}| = 1 \), we can rewrite this as:
   \[
   |P(e^{i\theta})| = |1 - ae^{i\theta} + a|.
   \]

5. **Simplify the expression:**
   Let \( w = e^{i\theta} \). Then,
   \[
   |P(w)| = |w^2 - aw + a|.
   \]
   Since \( |w| = 1 \), we have:
   \[
   |P(w)| = |1 - aw + a|.
   \]

6. **Find the minimum value:**
   To find the minimum value of \( |1 - aw + a| \) for \( |w| = 1 \), we need to consider the critical points and the boundary behavior. 

7. **Consider the critical points:**
   The critical points occur when the derivative of \( P(z) \) with respect to \( z \) is zero. However, since we are dealing with the magnitude, we need to consider the minimum value on the boundary \( |z| = 1 \).

8. **Evaluate at specific points:**
   Evaluate \( P(z) \) at specific points on the unit circle, such as \( z = 1 \) and \( z = -1 \):
   \[
   P(1) = 1 - a + a = 1,
   \]
   \[
   P(-1) = 1 + a + a = 1 + 2a.
   \]

9. **Analyze the range of \( a \):**
   Given \( a \in [2 + \sqrt{2}, 4] \), we need to check the minimum value of \( |P(z)| \) within this range.

10. **Conclusion:**
    Since \( |P(z)| \) is minimized when \( z = 1 \), the minimum value is:
    \[
    \inf_{\stackrel{z\in\mathbb{C}}{|z|\le 1}} \left| z^2-az+a \right| = 1.
    \]

The final answer is \( \boxed{1} \).