If `z_(1) = 2 - i , z_(2) = 1 + i , " find " |(z_(1) + z_(2) + 1)/( z_(1) -z_(2) + 1)|`

Answer

To solve the problem, we need to find the value of

\[ \left| \frac{z_1 + z_2 + 1}{z_1 - z_2 + 1} \right| \]

where \( z_1 = 2 - i \) and \( z_2 = 1 + i \).

### Step 1: Calculate \( z_1 + z_2 + 1 \)

First, we need to find \( z_1 + z_2 + 1 \):

\[ z_1 + z_2 = (2 - i) + (1 + i) = 2 + 1 + (-i + i) = 3 + 0i = 3 \]

Now, adding 1:

\[ z_1 + z_2 + 1 = 3 + 1 = 4 \]

### Step 2: Calculate \( z_1 - z_2 + 1 \)

Next, we calculate \( z_1 - z_2 + 1 \):

\[ z_1 - z_2 = (2 - i) - (1 + i) = 2 - 1 - i - i = 1 - 2i \]

Now, adding 1:

\[ z_1 - z_2 + 1 = (1 - 2i) + 1 = 2 - 2i \]

### Step 3: Substitute into the expression

Now we substitute these results into the expression:

\[ \left| \frac{z_1 + z_2 + 1}{z_1 - z_2 + 1} \right| = \left| \frac{4}{2 - 2i} \right| \]

### Step 4: Simplify the expression

To simplify \( \frac{4}{2 - 2i} \), we can multiply the numerator and the denominator by the conjugate of the denominator:

\[ \frac{4}{2 - 2i} \cdot \frac{2 + 2i}{2 + 2i} = \frac{4(2 + 2i)}{(2 - 2i)(2 + 2i)} \]

Calculating the denominator:

\[ (2 - 2i)(2 + 2i) = 2^2 - (2i)^2 = 4 - (-4) = 4 + 4 = 8 \]

Now, the numerator:

\[ 4(2 + 2i) = 8 + 8i \]

So we have:

\[ \frac{8 + 8i}{8} = 1 + i \]

### Step 5: Calculate the modulus

Now we find the modulus:

\[ \left| 1 + i \right| = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Final Answer

Thus, the final answer is:

\[ \left| \frac{z_1 + z_2 + 1}{z_1 - z_2 + 1} \right| = \sqrt{2} \]

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