There exists complex numbers $z=x+yi$ such that the point $(x, y)$ lies on the ellipse with equation $\frac{x^2}9+\frac{y^2}{16}=1$. If $\frac{z-1-i}{z-i}$ is real, compute $z$.

[i](Source: China National High School Mathematics League 2021, Zhejiang Province, Problem 3)[/i]
1. Let \( z = x + yi \) where \( x \) and \( y \) are real numbers. The point \((x, y)\) lies on the ellipse given by the equation:
   \[
   \frac{x^2}{9} + \frac{y^2}{16} = 1
   \]

2. We are given that \(\frac{z-1-i}{z-i}\) is real. This implies that the imaginary part of \(\frac{z-1-i}{z-i}\) must be zero. Let's express \( \frac{z-1-i}{z-i} \) in terms of \( x \) and \( y \):
   \[
   \frac{z-1-i}{z-i} = \frac{(x-1) + (y-1)i}{x + yi - i} = \frac{(x-1) + (y-1)i}{x + (y-1)i}
   \]

3. To make the denominator real, we multiply the numerator and the denominator by the conjugate of the denominator:
   \[
   \frac{(x-1) + (y-1)i}{x + (y-1)i} \cdot \frac{x - (y-1)i}{x - (y-1)i} = \frac{((x-1) + (y-1)i)(x - (y-1)i)}{(x + (y-1)i)(x - (y-1)i)}
   \]

4. Simplify the denominator:
   \[
   (x + (y-1)i)(x - (y-1)i) = x^2 - (y-1)^2i^2 = x^2 + (y-1)^2
   \]

5. Simplify the numerator:
   \[
   ((x-1) + (y-1)i)(x - (y-1)i) = (x-1)x - (x-1)(y-1)i + (y-1)ix - (y-1)^2i^2
   \]
   \[
   = x(x-1) - (y-1)^2 + i((y-1)x - (x-1)(y-1))
   \]
   \[
   = x^2 - x - (y-1)^2 + i((y-1)x - (x-1)(y-1))
   \]
   \[
   = x^2 - x - (y-1)^2 + i(y-1)(x - (x-1))
   \]
   \[
   = x^2 - x - (y-1)^2 + i(y-1)
   \]

6. For the fraction to be real, the imaginary part must be zero:
   \[
   (y-1) = 0 \implies y = 1
   \]

7. Substitute \( y = 1 \) into the ellipse equation:
   \[
   \frac{x^2}{9} + \frac{1^2}{16} = 1
   \]
   \[
   \frac{x^2}{9} + \frac{1}{16} = 1
   \]
   \[
   \frac{x^2}{9} = 1 - \frac{1}{16} = \frac{15}{16}
   \]
   \[
   x^2 = 9 \cdot \frac{15}{16} = \frac{135}{16}
   \]
   \[
   x = \pm \sqrt{\frac{135}{16}} = \pm \frac{\sqrt{135}}{4} = \pm \frac{3\sqrt{15}}{4}
   \]

8. Therefore, the complex numbers \( z \) are:
   \[
   z = \pm \frac{3\sqrt{15}}{4} + i
   \]

The final answer is \( \boxed{\pm \frac{3\sqrt{15}}{4} + i} \)