What is the modulus of `( 1+2i)/( 1- ( 1+i)^(2))` equal to ?

Options

A. A)5
B. B)4
C. C)3
D. D)`1`

Answer

To find the modulus of the expression \((1 + 2i)/(1 - (1 + i)^2)\), we can follow these steps:

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

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

Now, substituting this back into the denominator:

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

### Step 2: Rewrite the expression
Now we can rewrite the expression:

\[ \frac{1 + 2i}{1 - 2i} \]

### Step 3: Multiply by the conjugate
To simplify the fraction, we multiply the numerator and the denominator by the conjugate of the denominator, which is \(1 + 2i\):

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

### Step 4: Calculate the denominator
Now, calculate the denominator:

\[ (1 - 2i)(1 + 2i) = 1^2 - (2i)^2 = 1 - 4(-1) = 1 + 4 = 5 \]

### Step 5: Calculate the numerator
Now calculate the numerator:

\[ (1 + 2i)(1 + 2i) = 1^2 + 2 \cdot 1 \cdot 2i + (2i)^2 = 1 + 4i - 4 = -3 + 4i \]

### Step 6: Combine results
Now we can combine the results:

\[ \frac{-3 + 4i}{5} = -\frac{3}{5} + \frac{4}{5}i \]

### Step 7: Find the modulus
The modulus of a complex number \(z = x + yi\) is given by:

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

In our case, \(x = -\frac{3}{5}\) and \(y = \frac{4}{5}\):

\[ |z| = \sqrt{\left(-\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2} = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{\frac{25}{25}} = \sqrt{1} = 1 \]

### Final Answer
Thus, the modulus of \(\frac{1 + 2i}{1 - (1 + i)^2}\) is equal to **1**.

---