What is modulus of `(1)/(1+3i)-(1)/(1-3i)` ?

Options

A. `3/5`
B. `9/25`
C. `3/25`
D. `5/3`

Explanation

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

### Step 1: Find a common denominator
We need to combine the two fractions. The common denominator will be the product of the two denominators:
\[ (1 + 3i)(1 - 3i) \]
So we rewrite the expression:
\[ \frac{1}{1+3i} - \frac{1}{1-3i} = \frac{(1-3i) - (1+3i)}{(1+3i)(1-3i)} \]

### Step 2: Simplify the numerator
Now simplify the numerator:
\[ (1 - 3i) - (1 + 3i) = 1 - 3i - 1 - 3i = -6i \]
So the expression becomes:
\[ \frac{-6i}{(1+3i)(1-3i)} \]

### Step 3: Calculate the denominator
Now we calculate the denominator:
\[ (1 + 3i)(1 - 3i) = 1^2 - (3i)^2 = 1 - 9(-1) = 1 + 9 = 10 \]
Thus, we have:
\[ \frac{-6i}{10} = -\frac{3i}{5} \]

### Step 4: Find the modulus
The modulus of a complex number \(z = a + bi\) is given by:
\[ |z| = \sqrt{a^2 + b^2} \]
In our case, \(z = 0 - \frac{3i}{5}\), where the real part \(a = 0\) and the imaginary part \(b = -\frac{3}{5}\). Thus, we compute:
\[ |z| = \sqrt{0^2 + \left(-\frac{3}{5}\right)^2} = \sqrt{0 + \frac{9}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

### Final Answer
The modulus of the expression \(\frac{1}{1+3i} - \frac{1}{1-3i}\) is:
\[ \frac{3}{5} \]

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