Find the modulus of the following complex numbers
`(2+3i)/(3+2i)`

Answer

To find the modulus of the complex number \(\frac{2 + 3i}{3 + 2i}\), we will follow these steps:

### Step 1: Rationalize the denominator
To simplify the expression, we multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of \(3 + 2i\) is \(3 - 2i\).

\[ \frac{2 + 3i}{3 + 2i} \cdot \frac{3 - 2i}{3 - 2i} = \frac{(2 + 3i)(3 - 2i)}{(3 + 2i)(3 - 2i)} \]

### Step 2: Calculate the denominator
Using the identity \( (a + b)(a - b) = a^2 - b^2 \):

\[ (3 + 2i)(3 - 2i) = 3^2 - (2i)^2 = 9 - 4(-1) = 9 + 4 = 13 \]

### Step 3: Calculate the numerator
Now we expand the numerator:

\[ (2 + 3i)(3 - 2i) = 2 \cdot 3 + 2 \cdot (-2i) + 3i \cdot 3 + 3i \cdot (-2i) \]

Calculating each term:

- \(2 \cdot 3 = 6\)
- \(2 \cdot (-2i) = -4i\)
- \(3i \cdot 3 = 9i\)
- \(3i \cdot (-2i) = -6i^2 = 6\) (since \(i^2 = -1\))

Now combine these:

\[ 6 - 4i + 9i + 6 = (6 + 6) + (-4i + 9i) = 12 + 5i \]

### Step 4: Combine the results
Now we can write the expression as:

\[ \frac{12 + 5i}{13} \]

This can be separated into real and imaginary parts:

\[ \frac{12}{13} + \frac{5}{13}i \]

### Step 5: Find the modulus
The modulus of a complex number \(a + bi\) is given by:

\[ |z| = \sqrt{a^2 + b^2} \]

Here, \(a = \frac{12}{13}\) and \(b = \frac{5}{13}\):

\[ |z| = \sqrt{\left(\frac{12}{13}\right)^2 + \left(\frac{5}{13}\right)^2} \]

Calculating each square:

\[ |z| = \sqrt{\frac{144}{169} + \frac{25}{169}} = \sqrt{\frac{144 + 25}{169}} = \sqrt{\frac{169}{169}} = \sqrt{1} = 1 \]

### Conclusion
Thus, the modulus of the complex number \(\frac{2 + 3i}{3 + 2i}\) is:

\[ \boxed{1} \]