The modulus and principle argument of the complex number `(1+2i)/(1-(1-i)^(2)` are respectively

Options

A. `1,0`
B. `1,1`
C. `2,0`
D. `2,1`

Explanation

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

### Step 1: Simplify the Complex Number
We start with the complex number:
\[ z = \frac{1 + 2i}{1 - (1 - i)^2} \]

First, we need to simplify the denominator:
\[ (1 - i)^2 = 1^2 - 2(1)(i) + i^2 = 1 - 2i - 1 = -2i \]
So, the denominator becomes:
\[ 1 - (1 - i)^2 = 1 - (-2i) = 1 + 2i \]

Now we can rewrite \(z\):
\[ z = \frac{1 + 2i}{1 + 2i} \]

### Step 2: Simplify Further
Since the numerator and denominator are the same, we have:
\[ z = 1 \]

### Step 3: Find the Modulus
The modulus of a complex number \(z = a + bi\) is given by:
\[ |z| = \sqrt{a^2 + b^2} \]
For \(z = 1 + 0i\):
- Real part \(a = 1\)
- Imaginary part \(b = 0\)

Thus, the modulus is:
\[ |z| = \sqrt{1^2 + 0^2} = \sqrt{1} = 1 \]

### Step 4: Find the Principal Argument
The principal argument \(\theta\) of a complex number is given by:
\[ \theta = \tan^{-1}\left(\frac{b}{a}\right) \]
For \(z = 1 + 0i\):
\[ \theta = \tan^{-1}\left(\frac{0}{1}\right) = \tan^{-1}(0) = 0 \]

### Final Answer
Thus, the modulus and principal argument of the complex number are:
\[ \text{Modulus} = 1, \quad \text{Principal Argument} = 0 \]