Find the modulus and amplitude of the following complex numbers and hence express them into polar form
`((1+i)(2+i))/((3+i))`

Answer

To solve the problem of finding the modulus and amplitude of the complex number \(\frac{(1+i)(2+i)}{(3+i)}\) and expressing it in polar form, we will follow these steps:

### Step 1: Multiply the Numerator
We start with the expression:

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

First, we multiply the terms in the numerator:

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

Since \(i^2 = -1\), we can simplify this:

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

### Step 2: Rewrite the Expression
Now we rewrite \(z\):

\[ z = \frac{1 + 3i}{3 + i} \]

### Step 3: Multiply by the Conjugate
To simplify the division, we multiply the numerator and the denominator by the conjugate of the denominator:

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

Calculating the denominator:

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

Now for the numerator:

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

Again, since \(i^2 = -1\):

\[ = 3 - i + 9i + 3 = 6 + 8i \]

### Step 4: Final Expression
Now we can express \(z\):

\[ z = \frac{6 + 8i}{10} = \frac{6}{10} + \frac{8}{10}i = \frac{3}{5} + \frac{4}{5}i \]

### Step 5: Find the Modulus
The modulus of \(z\) is given by:

\[ |z| = \sqrt{\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2} \]

Calculating this:

\[ = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{\frac{25}{25}} = \sqrt{1} = 1 \]

### Step 6: Find the Amplitude
The amplitude (or argument) \(\theta\) can be found using:

\[ \tan \theta = \frac{\text{Imaginary part}}{\text{Real part}} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \]

Thus,

\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]

### Step 7: Express in Polar Form
The polar form of \(z\) is given by:

\[ z = r(\cos \theta + i \sin \theta) \]

Substituting the values we found:

\[ z = 1 \left( \cos \left(\tan^{-1}\left(\frac{4}{3}\right)\right) + i \sin \left(\tan^{-1}\left(\frac{4}{3}\right)\right) \right) \]

Thus, the polar form is:

\[ z = \cos \left(\tan^{-1}\left(\frac{4}{3}\right)\right) + i \sin \left(\tan^{-1}\left(\frac{4}{3}\right)\right) \]

### Summary of Results
- **Modulus**: \( |z| = 1 \)
- **Amplitude**: \( \theta = \tan^{-1}\left(\frac{4}{3}\right) \)
- **Polar Form**: \( z = \cos \left(\tan^{-1}\left(\frac{4}{3}\right)\right) + i \sin \left(\tan^{-1}\left(\frac{4}{3}\right)\right) \)