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 value of \( \left| \frac{z_1 + z_2 + 1}{z_1 - z_2 + i} \right| \) where \( z_1 = 2 - i \) and \( z_2 = 1 + i \).

### Step-by-Step Solution:

**Step 1: Calculate \( z_1 + z_2 + 1 \)**
We start by adding \( z_1 \) and \( z_2 \) and then adding 1 to the result.

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

Now, adding 1:

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

**Step 2: Calculate \( z_1 - z_2 + i \)**
Next, we need to compute \( z_1 - z_2 + i \).

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

Now, adding \( i \):

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

**Step 3: Form the fraction**
Now we can form the fraction:

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

**Step 4: Find the modulus of the fraction**
To find the modulus, we use the property \( | \frac{a}{b} | = \frac{|a|}{|b|} \):

\[ |4| = 4 \]

Now, we need to find \( |1 - i| \):

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

So, the modulus of the fraction is:

\[ \left| \frac{4}{1 - i} \right| = \frac{|4|}{|1 - i|} = \frac{4}{\sqrt{2}} \]

**Step 5: Rationalize the denominator**
To simplify \( \frac{4}{\sqrt{2}} \):

\[ \frac{4}{\sqrt{2}} = \frac{4 \cdot \sqrt{2}}{2} = 2\sqrt{2} \]

### Final Answer:
Thus, the value of \( \left| \frac{z_1 + z_2 + 1}{z_1 - z_2 + i} \right| \) is \( 2\sqrt{2} \).

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