A complex number is given by `z=(1+2i)/(1-(1-i)^(2))`
What is the modulus of z ?

Options

A. 4
B. 2
C. 1
D. `1/2`

Explanation

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

### Step 1: Simplify the denominator
We need to simplify \( 1 - (1 - i)^2 \).

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

2. Substitute this back into the expression for \( z \):
\[ 1 - (1 - i)^2 = 1 - (-2i) = 1 + 2i \]

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

### Step 3: Simplify \( z \)
Since the numerator and denominator are the same:
\[ z = 1 \]

### Step 4: Find the modulus of \( z \)
The modulus of a complex number \( z = a + bi \) is given by:
\[ |z| = \sqrt{a^2 + b^2} \]
In our case, \( z = 1 + 0i \), so \( a = 1 \) and \( b = 0 \):
\[ |z| = \sqrt{1^2 + 0^2} = \sqrt{1} = 1 \]

### Final Answer
The modulus of \( z \) is \( 1 \).

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