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, we need to find the modulus of the expression \(\frac{z_1 + z_2 + 1}{z_1 - z_2 + i}\), where \(z_1 = 2 - i\) and \(z_2 = 1 + i\).

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

First, we find \(z_1 + z_2 + 1\):
\[ z_1 + z_2 = (2 - i) + (1 + i) = 2 + 1 - i + i = 3 \]
Now, adding 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, adding \(i\):
\[ z_1 - z_2 + i = (1 - 2i) + i = 1 - 2i + i = 1 - i \]

### Step 3: Form the expression

Now we can form the expression:
\[ \frac{z_1 + z_2 + 1}{z_1 - z_2 + i} = \frac{4}{1 - i} \]

### Step 4: Multiply by the conjugate

To simplify this expression, we multiply the numerator and the denominator by the conjugate of the denominator:
\[ \text{Conjugate of } (1 - i) = (1 + i) \]
Thus, we have:
\[ \frac{4}{1 - i} \cdot \frac{1 + i}{1 + i} = \frac{4(1 + i)}{(1 - i)(1 + i)} \]

### Step 5: Simplify the denominator

Calculating the denominator:
\[ (1 - i)(1 + i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2 \]
So, we have:
\[ \frac{4(1 + i)}{2} = \frac{4}{2}(1 + i) = 2(1 + i) = 2 + 2i \]

### Step 6: Find the modulus

Now, we find the modulus of \(2 + 2i\):
\[ |2 + 2i| = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

### Final Answer

Thus, the final answer is:
\[ \left| \frac{z_1 + z_2 + 1}{z_1 - z_2 + i} \right| = 2\sqrt{2} \]