Find the modulus and argument of the complex number `(2 + i)/(4i + (1+ i)^(2))`

Answer

To find the modulus and argument of the complex number \(\frac{2 + i}{4i + (1 + i)^2}\), we will follow these steps:

### Step 1: Simplify the Denominator
First, we need to simplify the denominator \(4i + (1 + i)^2\).

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

Now substituting this back into the denominator:

\[ 4i + (1 + i)^2 = 4i + 2i = 6i \]

### Step 2: Rewrite the Complex Number
Now we can rewrite the complex number:

\[ \frac{2 + i}{6i} \]

### Step 3: Separate into Real and Imaginary Parts
We can separate the numerator:

\[ \frac{2 + i}{6i} = \frac{2}{6i} + \frac{i}{6i} = \frac{2}{6i} + \frac{1}{6} \]

To simplify \(\frac{2}{6i}\):

\[ \frac{2}{6i} = \frac{2}{6} \cdot \frac{-i}{-i} = \frac{-2i}{6} = -\frac{1}{3}i \]

So we have:

\[ -\frac{1}{3}i + \frac{1}{6} = \frac{1}{6} - \frac{1}{3}i \]

### Step 4: Identify Real and Imaginary Parts
Now, we can identify the real and imaginary parts:

\[ z = \frac{1}{6} - \frac{1}{3}i \]

Here, \(a = \frac{1}{6}\) and \(b = -\frac{1}{3}\).

### Step 5: Calculate the Modulus
The modulus \(r\) of a complex number \(z = a + bi\) is given by:

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

Calculating:

\[ r = \sqrt{\left(\frac{1}{6}\right)^2 + \left(-\frac{1}{3}\right)^2} = \sqrt{\frac{1}{36} + \frac{1}{9}} = \sqrt{\frac{1}{36} + \frac{4}{36}} = \sqrt{\frac{5}{36}} = \frac{\sqrt{5}}{6} \]

### Step 6: Calculate the Argument
The argument \(\theta\) can be found using:

\[ \tan \theta = \frac{b}{a} = \frac{-\frac{1}{3}}{\frac{1}{6}} = -2 \]

Now we find \(\theta\):

\[ \theta = \tan^{-1}(-2) \]

Since \(a > 0\) and \(b < 0\), the complex number lies in the fourth quadrant. Thus, the angle will be negative:

\[ \theta \approx -63.43^\circ \]

### Final Result
The modulus and argument of the complex number \(\frac{2 + i}{4i + (1 + i)^2}\) are:

\[ \text{Modulus} = \frac{\sqrt{5}}{6}, \quad \text{Argument} = -63.43^\circ \]