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

Options

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

Answer

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 denominator
First, we need to simplify the denominator \(1 - (1 - i)^2\).

\[ (1 - i)^2 = 1^2 - 2(1)(i) + (i)^2 = 1 - 2i + (-1) = 0 - 2i = -2i \]

Now substituting this back into the denominator:

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

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

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

### Step 3: Simplify the complex number
Since the numerator and denominator are the same, we simplify this to:

\[ 1 \]

### Step 4: Find the modulus
The modulus of a complex number \(z = a + bi\) is given by:

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

For \(z = 1\):

\[ |1| = \sqrt{1^2 + 0^2} = \sqrt{1} = 1 \]

### Step 5: Find the principal argument
The principal argument \(\theta\) of a complex number is given by:

\[ \theta = \tan^{-1}\left(\frac{b}{a}\right) \]

For \(z = 1\) (where \(a = 1\) and \(b = 0\)):

\[ \theta = \tan^{-1}\left(\frac{0}{1}\right) = \tan^{-1}(0) = 0 \]

### Final Result
Thus, the modulus and principal argument of the complex number \(\frac{1 + 2i}{1 - (1 - i)^2}\) are:

- Modulus: \(1\)
- Principal Argument: \(0\)