Let `|z_(1)|=3, |z_(2)|=2 and z_(1)+z_(2)+z_(3)=3+4i`. If the real part of `(z_(1)bar(z_(2))+z_(2)bar(z_(3))+z_(3)bar(z_(1)))` is equal to 4, then `|z_(3)|` is equal to (where, `i^(2)=-1`)

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem step-by-step, we will follow the given information and apply mathematical principles related to complex numbers.

### Step 1: Understand the given information
We know:
- \( |z_1| = 3 \)
- \( |z_2| = 2 \)
- \( z_1 + z_2 + z_3 = 3 + 4i \)
- The real part of \( (z_1 \bar{z_2} + z_2 \bar{z_3} + z_3 \bar{z_1}) = 4 \)

### Step 2: Calculate the modulus of the sum
We can calculate the modulus of the sum \( z_1 + z_2 + z_3 \):
\[ |z_1 + z_2 + z_3| = |3 + 4i| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

### Step 3: Use the property of modulus
Since \( |z_1 + z_2 + z_3| \leq |z_1| + |z_2| + |z_3| \), we can write:
\[ 5 \leq |z_1| + |z_2| + |z_3| \]
Substituting the known values:
\[ 5 \leq 3 + 2 + |z_3| \implies |z_3| \geq 0 \]
This does not give us a specific value yet.

### Step 4: Square the modulus equation
Now we will square the modulus equation:
\[ |z_1 + z_2 + z_3|^2 = |z_1|^2 + |z_2|^2 + |z_3|^2 + 2 \text{Re}(z_1 \bar{z_2} + z_2 \bar{z_3} + z_3 \bar{z_1}) \]
We know:
\[ |z_1|^2 = 9, \quad |z_2|^2 = 4, \quad |z_3|^2 = |z_3|^2 \]
Thus:
\[ 25 = 9 + 4 + |z_3|^2 + 2 \text{Re}(z_1 \bar{z_2} + z_2 \bar{z_3} + z_3 \bar{z_1}) \]

### Step 5: Substitute the real part value
We know that the real part \( \text{Re}(z_1 \bar{z_2} + z_2 \bar{z_3} + z_3 \bar{z_1}) = 4 \):
\[ 25 = 9 + 4 + |z_3|^2 + 2 \cdot 4 \]
This simplifies to:
\[ 25 = 9 + 4 + |z_3|^2 + 8 \]
\[ 25 = 21 + |z_3|^2 \]

### Step 6: Solve for \( |z_3|^2 \)
Rearranging gives:
\[ |z_3|^2 = 25 - 21 = 4 \]
Taking the square root:
\[ |z_3| = 2 \]

### Conclusion
Thus, the value of \( |z_3| \) is \( 2 \).