If `|z_1=1,|z_2|=2,|z_3|=3 and |z_1+z_2+z_3|=1, then |9z_1z_2+4z_3z_1+z_2z_3|` is equal to

Options

A. 3
B. 6
C. 36
D. 216

Answer

To solve the problem step by step, we need to find the value of \( |9z_1z_2 + 4z_3z_1 + z_2z_3| \) given the conditions \( |z_1| = 1 \), \( |z_2| = 2 \), \( |z_3| = 3 \), and \( |z_1 + z_2 + z_3| = 1 \).

### Step 1: Factor out the common terms
We can factor out \( z_1 z_2 z_3 \) from the expression:
\[ |9z_1z_2 + 4z_3z_1 + z_2z_3| = |z_1 z_2 z_3| \cdot |9/z_3 + 4/z_2 + 1/z_1| \]

### Step 2: Calculate the modulus of \( z_1 z_2 z_3 \)
Using the properties of modulus:
\[ |z_1 z_2 z_3| = |z_1| \cdot |z_2| \cdot |z_3| = 1 \cdot 2 \cdot 3 = 6 \]

### Step 3: Evaluate \( |9/z_3 + 4/z_2 + 1/z_1| \)
Now we need to evaluate the term \( |9/z_3 + 4/z_2 + 1/z_1| \). We will rewrite each term using the conjugates:
\[ \frac{9}{z_3} = \frac{9 \cdot \overline{z_3}}{|z_3|^2}, \quad \frac{4}{z_2} = \frac{4 \cdot \overline{z_2}}{|z_2|^2}, \quad \frac{1}{z_1} = \frac{1 \cdot \overline{z_1}}{|z_1|^2} \]
Substituting the moduli:
\[ |9/z_3 + 4/z_2 + 1/z_1| = \left| \frac{9 \cdot \overline{z_3}}{9} + \frac{4 \cdot \overline{z_2}}{4} + \frac{\overline{z_1}}{1} \right| = |\overline{z_3} + \overline{z_2} + \overline{z_1}| \]

### Step 4: Use the property of conjugates
Using the property of conjugates:
\[ |\overline{z_1} + \overline{z_2} + \overline{z_3}| = |z_1 + z_2 + z_3| \]
Given \( |z_1 + z_2 + z_3| = 1 \), we have:
\[ |\overline{z_1} + \overline{z_2} + \overline{z_3}| = 1 \]

### Step 5: Combine the results
Now we can combine our results:
\[ |9z_1z_2 + 4z_3z_1 + z_2z_3| = |z_1 z_2 z_3| \cdot |9/z_3 + 4/z_2 + 1/z_1| = 6 \cdot 1 = 6 \]

### Final Answer
Thus, the value of \( |9z_1z_2 + 4z_3z_1 + z_2z_3| \) is \( \boxed{6} \).