Let $z$ be a complex number such that $|z| = 1$ and $|z-1.45|=1.05$. Compute the real part of $z$.
1. Let \( z = a + bi \). Given that \( |z| = 1 \), we have:
   \[
   |z| = \sqrt{a^2 + b^2} = 1 \implies a^2 + b^2 = 1
   \]

2. We are also given that \( |z - 1.45| = 1.05 \). This translates to:
   \[
   |(a + bi) - 1.45| = 1.05 \implies |(a - 1.45) + bi| = 1.05
   \]
   Therefore:
   \[
   \sqrt{(a - 1.45)^2 + b^2} = 1.05 \implies (a - 1.45)^2 + b^2 = 1.05^2 = 1.1025
   \]

3. We now have two equations:
   \[
   a^2 + b^2 = 1 \quad \text{(Equation 1)}
   \]
   \[
   (a - 1.45)^2 + b^2 = 1.1025 \quad \text{(Equation 2)}
   \]

4. Subtract Equation 1 from Equation 2:
   \[
   (a - 1.45)^2 + b^2 - (a^2 + b^2) = 1.1025 - 1
   \]
   Simplifying, we get:
   \[
   (a - 1.45)^2 - a^2 = 0.1025
   \]

5. Expand and simplify the left-hand side:
   \[
   (a - 1.45)^2 = a^2 - 2 \cdot 1.45 \cdot a + 1.45^2
   \]
   \[
   a^2 - 2.9a + 2.1025 - a^2 = 0.1025
   \]
   \[
   -2.9a + 2.1025 = 0.1025
   \]

6. Solve for \( a \):
   \[
   -2.9a + 2.1025 = 0.1025
   \]
   \[
   -2.9a = 0.1025 - 2.1025
   \]
   \[
   -2.9a = -2
   \]
   \[
   a = \frac{2}{2.9} = \frac{20}{29}
   \]

7. Therefore, the real part of \( z \) is:
   \[
   \boxed{\frac{20}{29}}
   \]