Find the modulus and amplitude of ` ( 2+ i)/( 4i +(1+i) ^(2))`

Answer

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

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

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

Now substitute this back into the denominator:

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

### Step 2: Rewrite the complex number
Now we can rewrite \( z \):

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

### Step 3: Separate the real and imaginary parts
We can separate the terms in the fraction:

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

### Step 4: Simplify the first term
To simplify \( \frac{2}{6i} \), we multiply the numerator and denominator by \( -i \):

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

Now substituting this back into \( z \):

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

### Step 5: Combine the terms
Now we can express \( z \) in the standard form \( x + iy \):

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

### Step 6: Identify the real and imaginary parts
From \( z = \frac{1}{6} - \frac{1}{3}i \), we identify:

- Real part \( x = \frac{1}{6} \)
- Imaginary part \( y = -\frac{1}{3} \)

### Step 7: Calculate the modulus
The modulus \( |z| \) is given by:

\[ |z| = \sqrt{x^2 + y^2} = \sqrt{\left(\frac{1}{6}\right)^2 + \left(-\frac{1}{3}\right)^2} \]

Calculating:

\[ |z| = \sqrt{\frac{1}{36} + \frac{1}{9}} = \sqrt{\frac{1}{36} + \frac{4}{36}} = \sqrt{\frac{5}{36}} = \frac{\sqrt{5}}{6} \]

### Step 8: Calculate the amplitude
The amplitude \( \theta \) is given by:

\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{-\frac{1}{3}}{\frac{1}{6}}\right) = \tan^{-1}\left(-2\right) \]

### Final Result
Thus, the modulus and amplitude of \( z \) are:

- Modulus: \( |z| = \frac{\sqrt{5}}{6} \)
- Amplitude: \( \theta = \tan^{-1}(-2) \)