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

Answer

To find the modulus and amplitude 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 \(4i + (1+i)^2\).

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

Now substituting back into the denominator:
\[ 4i + (1+i)^2 = 4i + 2i = 6i \]

### Step 2: Rewrite the Expression
Now we can rewrite the expression:
\[ \frac{2+i}{6i} \]

### Step 3: Multiply by the Conjugate
To simplify further, we can multiply the numerator and the denominator by \(i\) (the conjugate of \(6i\)):
\[ \frac{(2+i)i}{6i^2} = \frac{2i + i^2}{6(-1)} = \frac{2i - 1}{-6} = \frac{-1 + 2i}{6} \]

### Step 4: Separate Real and Imaginary Parts
Now we can express this as:
\[ \frac{-1}{6} + \frac{2}{6}i = -\frac{1}{6} + \frac{1}{3}i \]

### Step 5: Find the Modulus
The modulus of a complex number \(a + bi\) is given by:
\[ |z| = \sqrt{a^2 + b^2} \]
Here, \(a = -\frac{1}{6}\) and \(b = \frac{1}{3}\):
\[ |z| = \sqrt{\left(-\frac{1}{6}\right)^2 + \left(\frac{1}{3}\right)^2} = \sqrt{\frac{1}{36} + \frac{1}{9}} = \sqrt{\frac{1}{36} + \frac{4}{36}} = \sqrt{\frac{5}{36}} = \frac{\sqrt{5}}{6} \]

### Step 6: Find the Amplitude
The amplitude (or argument) \(\theta\) of a complex number is given by:
\[ \theta = \tan^{-1}\left(\frac{b}{a}\right) \]
Substituting the values:
\[ \theta = \tan^{-1}\left(\frac{\frac{1}{3}}{-\frac{1}{6}}\right) = \tan^{-1}\left(-2\right) \]

Since the complex number is in the second quadrant (negative real part and positive imaginary part), we can express this as:
\[ \theta = \pi + \tan^{-1}(-2) = -\tan^{-1}(2) \]

### Final Result
Thus, the modulus and amplitude of the complex number \(\frac{2+i}{4i+(1+i)^2}\) are:
\[ \text{Modulus} = \frac{\sqrt{5}}{6}, \quad \text{Amplitude} = -\tan^{-1}(2) \]