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

Options

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

Answer

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

### Step 1: Combine the fractions
To combine the two fractions, we need a common denominator. The common denominator will be \((1 + 3i)(1 - 3i)\).

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

### Step 2: Simplify the numerator
Now, simplify the numerator:

\[ (1 - 3i) - (1 + 3i) = 1 - 3i - 1 - 3i = -6i \]

### Step 3: Simplify the denominator
Next, simplify the denominator using the difference of squares:

\[ (1 + 3i)(1 - 3i) = 1^2 - (3i)^2 = 1 - 9(-1) = 1 + 9 = 10 \]

### Step 4: Combine the results
Now we can combine the results from the numerator and denominator:

\[ \frac{-6i}{10} = -\frac{3i}{5} \]

### Step 5: Find the modulus
The modulus of a complex number \(a + bi\) is given by \(\sqrt{a^2 + b^2}\). In our case, we have:

\[ -\frac{3i}{5} = 0 - \frac{3}{5}i \]

Here, \(a = 0\) and \(b = -\frac{3}{5}\). Thus, the modulus is:

\[ \sqrt{0^2 + \left(-\frac{3}{5}\right)^2} = \sqrt{0 + \frac{9}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

### Final Answer
The modulus of \(\frac{1}{1+3i} - \frac{1}{1-3i}\) is \(\frac{3}{5}\).

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