Find the modulus and argument of the complex number `(1+2i)/(1-3i)`.

Explanation

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

### Step 1: Multiply by the Conjugate
We start by multiplying the numerator and the denominator by the conjugate of the denominator. The conjugate of \(1 - 3i\) is \(1 + 3i\).

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

### Step 2: Simplify the Denominator
The denominator can be simplified using the difference of squares formula:

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

### Step 3: Simplify the Numerator
Now, we simplify the numerator:

\[ (1+2i)(1+3i) = 1 \cdot 1 + 1 \cdot 3i + 2i \cdot 1 + 2i \cdot 3i = 1 + 3i + 2i + 6i^2 \]

Since \(i^2 = -1\), we have:

\[ 6i^2 = 6(-1) = -6 \]

Thus, the numerator becomes:

\[ 1 + 3i + 2i - 6 = -5 + 5i \]

### Step 4: Combine the Results
Now we can combine the results:

\[ \frac{-5 + 5i}{10} = -\frac{1}{2} + \frac{1}{2}i \]

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

\[ |z| = \sqrt{x^2 + y^2} \]

In our case, \(x = -\frac{1}{2}\) and \(y = \frac{1}{2}\):

\[ |z| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{2}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

### Step 6: Find the Argument
The argument \(\theta\) of a complex number can be found using:

\[ \tan \theta = \frac{y}{x} \]

Here, \(y = \frac{1}{2}\) and \(x = -\frac{1}{2}\):

\[ \tan \theta = \frac{\frac{1}{2}}{-\frac{1}{2}} = -1 \]

The angle whose tangent is \(-1\) is \(-\frac{\pi}{4}\) or \(\frac{3\pi}{4}\) (since we are in the second quadrant where \(x\) is negative and \(y\) is positive).

Thus, the argument is:

\[ \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \]

### Final Result
The modulus and argument of the complex number \(\frac{1+2i}{1-3i}\) are:

\[ \text{Modulus} = \frac{1}{\sqrt{2}}, \quad \text{Argument} = \frac{3\pi}{4} \]