If z =`(1)/((2-3i)^(2))` then |z| is equal to

Options

A. `(1)/(13)`
B. `(1)/(12)`
C. `(1)/(5)`
D. 13

Answer

To find the modulus of \( z = \frac{1}{(2 - 3i)^2} \), we will follow these steps:

### Step 1: Calculate \( (2 - 3i)^2 \)

Using the formula for the square of a binomial, \( (a - b)^2 = a^2 - 2ab + b^2 \):

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

Calculating each term:

- \( 2^2 = 4 \)
- \( -2 \cdot 2 \cdot 3i = -12i \)
- \( (3i)^2 = 9i^2 = 9(-1) = -9 \)

Putting it all together:

\[ (2 - 3i)^2 = 4 - 12i - 9 = -5 - 12i \]

### Step 2: Substitute back into \( z \)

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

\[ z = \frac{1}{-5 - 12i} \]

### Step 3: Multiply by the conjugate

To simplify \( z \), we multiply the numerator and the denominator by the conjugate of the denominator:

\[ z = \frac{1 \cdot (-5 + 12i)}{(-5 - 12i)(-5 + 12i)} \]

Calculating the denominator:

\[ (-5 - 12i)(-5 + 12i) = (-5)^2 - (12i)^2 = 25 - 144(-1) = 25 + 144 = 169 \]

So we have:

\[ z = \frac{-5 + 12i}{169} \]

### Step 4: Find the modulus of \( z \)

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

In our case:

\[ z = \frac{-5}{169} + \frac{12i}{169} \]

Thus, \( a = -\frac{5}{169} \) and \( b = \frac{12}{169} \).

Calculating the modulus:

\[ |z| = \sqrt{\left(-\frac{5}{169}\right)^2 + \left(\frac{12}{169}\right)^2} \]

Calculating each square:

\[ \left(-\frac{5}{169}\right)^2 = \frac{25}{28561} \]
\[ \left(\frac{12}{169}\right)^2 = \frac{144}{28561} \]

Adding these:

\[ |z| = \sqrt{\frac{25 + 144}{28561}} = \sqrt{\frac{169}{28561}} = \frac{\sqrt{169}}{\sqrt{28561}} = \frac{13}{169} \]

### Step 5: Final result

Thus, the modulus of \( z \) is:

\[ |z| = \frac{1}{13} \]