Examine the continuity of the function f(z) =( z^2-3z+4)÷(z^2 -1) at z=1-i

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) = 

Examine the continuity of the function f(z) = (z^2 - 3z + 4) ÷ (z^2 - 1) at z = 1 - i.

Explore the continuity of the complex function f(z) = (z^2 - 3z + 4) ÷ (z^2 - 1) at the point z = 1 - i. This detailed solution includes key concepts from complex analysis, such as rational functions and multiplication by complex conjugates. Suitable for undergraduate students studying complex analysis.

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