Finding the Modulus of the Given Complex Number z

The problem asks us to find the modulus of the complex number \(z\) given by the expression:

\(z = \frac{{1 + 2i}}{{1 - {{\left( {1 - i} \right)}^2}}}\)

To find the modulus \(|z|\), we first need to simplify the expression for \(z\). Let's start by simplifying the denominator.

Simplifying the Denominator

The denominator is \(1 - {{\left( {1 - i} \right)}^2}\). Let's first calculate \({\left( {1 - i} \right)}^2\).

Using the formula \({\left( {a - b} \right)}^2 = a^2 - 2ab + b^2\):

\({\left( {1 - i} \right)}^2 = 1^2 - 2(1)(i) + i^2\)

We know that \(i^2 = -1\). Substituting this value:

\({\left( {1 - i} \right)}^2 = 1 - 2i + (-1)\)

\({\left( {1 - i} \right)}^2 = 1 - 2i - 1\)

\({\left( {1 - i} \right)}^2 = -2i\)

Now, substitute this back into the denominator expression:

Denominator \(= 1 - {{\left( {1 - i} \right)}^2} = 1 - (-2i)\)

Denominator \(= 1 + 2i\)

Simplifying the Complex Number z

Now we can substitute the simplified denominator back into the expression for \(z\):

\(z = \frac{{1 + 2i}}{{1 + 2i}}\)

Since the numerator and the denominator are the same, the fraction simplifies to 1.

\(z = 1\)

The complex number \(z\) simplifies to \(1\). We can write this in the form \(a + bi\) as \(1 + 0i\).

Calculating the Modulus of z

The modulus of a complex number \(a + bi\) is given by the formula \(|a + bi| = \sqrt{a^2 + b^2}\).

For \(z = 1 + 0i\), we have \(a = 1\) and \(b = 0\).

The modulus of \(z\) is:

\(|z| = \sqrt{1^2 + 0^2}\)

\(|z| = \sqrt{1 + 0}\)

\(|z| = \sqrt{1}\)

\(|z| = 1\)

Thus, the modulus of the complex number \(z\) is 1.

Comparing with the given options, the correct modulus is 1.

Revision Table: Complex Number Modulus

Additional Information: Operations with Complex Numbers

Addition/Subtraction: Add or subtract the real and imaginary parts separately.
\((a + bi) \pm (c + di) = (a \pm c) + (b \pm d)i\)

Multiplication: Multiply complex numbers like binomials, remembering \(i^2 = -1\).
\((a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i\)

Division: Multiply the numerator and denominator by the conjugate of the denominator. The conjugate of \(c + di\) is \(c - di\).
\(\frac{{a + bi}}{{c + di}} = \frac{{(a + bi)(c - di)}}{{(c + di)(c - di)}} = \frac{{(ac + bd) + (bc - ad)i}}{{c^2 + d^2}}\)

Conjugate: The conjugate of \(a + bi\) is \(\overline{a + bi} = a - bi\).

In this problem, we used the property of complex number multiplication to expand \({\left( {1 - i} \right)}^2\) and the formula for the modulus of a complex number to find the final answer.