Find the modulus of `((3+2i) (1+i) (2+3i))/((3+4i) (4+5i))`

Answer

To find the modulus of the expression \(\frac{(3+2i)(1+i)(2+3i)}{(3+4i)(4+5i)}\), we will follow these steps:

### Step 1: Calculate the product in the numerator
Let \( z_1 = 3 + 2i \), \( z_2 = 1 + i \), and \( z_3 = 2 + 3i \).

We first calculate the product \( z_1 z_2 z_3 \).

1. Calculate \( z_1 z_2 \):
\[ z_1 z_2 = (3 + 2i)(1 + i) = 3 \cdot 1 + 3 \cdot i + 2i \cdot 1 + 2i \cdot i = 3 + 3i + 2i + 2i^2 \]
Since \( i^2 = -1 \):
\[ = 3 + 5i - 2 = 1 + 5i \]

2. Now calculate \( (1 + 5i)(2 + 3i) \):
\[ (1 + 5i)(2 + 3i) = 1 \cdot 2 + 1 \cdot 3i + 5i \cdot 2 + 5i \cdot 3i = 2 + 3i + 10i + 15i^2 \]
Again, using \( i^2 = -1 \):
\[ = 2 + 13i - 15 = -13 + 13i \]

So, the numerator is \( -13 + 13i \).

### Step 2: Calculate the product in the denominator
Let \( z_4 = 3 + 4i \) and \( z_5 = 4 + 5i \).

1. Calculate \( z_4 z_5 \):
\[ z_4 z_5 = (3 + 4i)(4 + 5i) = 3 \cdot 4 + 3 \cdot 5i + 4i \cdot 4 + 4i \cdot 5i = 12 + 15i + 16i + 20i^2 \]
Using \( i^2 = -1 \):
\[ = 12 + 31i - 20 = -8 + 31i \]

So, the denominator is \( -8 + 31i \).

### Step 3: Find the modulus of the numerator and denominator
The modulus of a complex number \( a + bi \) is given by \( \sqrt{a^2 + b^2} \).

1. Modulus of the numerator \( -13 + 13i \):
\[ |z_{numerator}| = \sqrt{(-13)^2 + (13)^2} = \sqrt{169 + 169} = \sqrt{338} \]

2. Modulus of the denominator \( -8 + 31i \):
\[ |z_{denominator}| = \sqrt{(-8)^2 + (31)^2} = \sqrt{64 + 961} = \sqrt{1025} \]

### Step 4: Find the modulus of the entire expression
The modulus of the fraction is given by:
\[ \left| \frac{z_{numerator}}{z_{denominator}} \right| = \frac{|z_{numerator}|}{|z_{denominator}|} = \frac{\sqrt{338}}{\sqrt{1025}} = \sqrt{\frac{338}{1025}} \]

### Step 5: Simplify the expression
To simplify \( \frac{338}{1025} \):
- Factor \( 338 = 2 \times 13^2 \)
- Factor \( 1025 = 5^2 \times 41 \)

Thus, we can write:
\[ \sqrt{\frac{338}{1025}} = \frac{13\sqrt{2}}{5\sqrt{41}} \]

### Final Answer
The modulus of the expression is:
\[ \frac{13\sqrt{2}}{5\sqrt{41}} \]