If `z_1=2-i ,z_2=1+i ,`find `|(z_1+z_2+1)/(z_1-z_2+i)|`

Explanation

To solve the problem \( | \frac{z_1 + z_2 + 1}{z_1 - z_2 + i} | \) where \( z_1 = 2 - i \) and \( z_2 = 1 + i \), we will follow these steps:

### 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 add 1:

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

### Step 2: Calculate \( z_1 - z_2 + i \)
Next, we calculate \( z_1 - z_2 + i \).

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

Now add \( i \):

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

### Step 3: Formulate the expression
Now we can write the expression we need to find the modulus of:

\[ \frac{z_1 + z_2 + 1}{z_1 - z_2 + i} = \frac{4}{1 - i} \]

### Step 4: Multiply by the conjugate
To simplify \( \frac{4}{1 - i} \), we multiply the numerator and the denominator by the conjugate of the denominator, which is \( 1 + i \):

\[ \frac{4(1 + i)}{(1 - i)(1 + i)} = \frac{4(1 + i)}{1^2 - (-i^2)} = \frac{4(1 + i)}{1 + 1} = \frac{4(1 + i)}{2} = 2(1 + i) \]

### Step 5: Find the modulus
Now we need to find the modulus of \( 2(1 + i) \):

\[ |2(1 + i)| = 2 |1 + i| \]

The modulus of \( 1 + i \) is given by:

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

Thus,

\[ |2(1 + i)| = 2 \sqrt{2} \]

### Final Answer
Therefore, the final answer is:

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