The modulus of the complex number `(1)/(2-3i)` is. . . .

Answer

To find the modulus of the complex number \( \frac{1}{2 - 3i} \), we will follow these steps:

### Step 1: Identify the complex number
Let \( z = \frac{1}{2 - 3i} \).

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

\[ z = \frac{1 \cdot (2 + 3i)}{(2 - 3i)(2 + 3i)} \]

### Step 3: Simplify the denominator
Now, calculate the denominator:

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

### Step 4: Simplify the numerator
The numerator becomes:

\[ 1 \cdot (2 + 3i) = 2 + 3i \]

### Step 5: Combine the results
Now, we can write \( z \) as:

\[ z = \frac{2 + 3i}{13} = \frac{2}{13} + \frac{3}{13}i \]

### Step 6: Identify the real and imaginary parts
From \( z = \frac{2}{13} + \frac{3}{13}i \), we identify:
- Real part \( a = \frac{2}{13} \)
- Imaginary part \( b = \frac{3}{13} \)

### Step 7: Calculate the modulus
The modulus of a complex number \( z = a + bi \) is given by:

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

Substituting the values of \( a \) and \( b \):

\[ |z| = \sqrt{\left(\frac{2}{13}\right)^2 + \left(\frac{3}{13}\right)^2} \]

### Step 8: Calculate the squares
Calculating the squares:

\[ \left(\frac{2}{13}\right)^2 = \frac{4}{169} \]
\[ \left(\frac{3}{13}\right)^2 = \frac{9}{169} \]

### Step 9: Add the squares
Now, add the squares:

\[ |z| = \sqrt{\frac{4}{169} + \frac{9}{169}} = \sqrt{\frac{13}{169}} = \sqrt{\frac{1}{13}} \]

### Step 10: Final result
Thus, the modulus of the complex number \( \frac{1}{2 - 3i} \) is:

\[ |z| = \frac{1}{\sqrt{13}} \]

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