To analyze the continuity of the function $f(z) = \frac{z^2 - 3z + 4}{z^2 - 1}$ at $z = 1 - i$, we need to check whether the function is well-defined and continuous at that point.

Step 1: Check if the function is defined at $z = 1 - i$

For a function to be continuous at a point $z_0$, it must first be defined at $z_0$. This means the denominator should not be zero at $z = 1 - i$.

We compute the value of the denominator $z^2 - 1$ at $z = 1 - i$:

$z^2 = (1 - i)^2 = 1^2 - 2(1)(i) + i^2 = 1 - 2i - 1 = -2i$ $z^2 - 1 = -2i - 1$

Thus, the denominator at $z = 1 - i$ is: $z^2 - 1 = -1 - 2i$

Since the denominator is not zero, the function is defined at $z = 1 - i$.

Step 2: Compute the value of the numerator at $z = 1 - i$

Now, calculate the value of the numerator $z^2 - 3z + 4$ at $z = 1 - i$:

We already know that $z^2 = -2i$. Now, let's calculate $-3z + 4$: $-3z = -3(1 - i) = -3 + 3i$ $z^2 - 3z + 4 = (-2i) + (-3 + 3i) + 4 = 1 + i$

Thus, the numerator at $z = 1 - i$ is: $z^2 - 3z + 4 = 1 + i$

Step 3: Determine the function's value at $z = 1 - i$

The function at $z = 1 - i$ is: $f(1 - i) = \frac{1 + i}{-1 - 2i}$

To simplify this, multiply both the numerator and denominator by the conjugate of the denominator: $f(1 - i) = \frac{(1 + i)(-1 + 2i)}{(-1 - 2i)(-1 + 2i)}$ The denominator becomes: [ (-1 - 2i)(-1 + 2i) = (-1)^2 - (2i)^2 = 1 - (-4) =