Let `z_1, z_2, …...z_n` be a sequence of complex numbers defined by `z_1=-1-i and z_(n+1) = z_(n)^(2) -i` for all `n ge 1`, then `|z_(2013)|` is equal to

Options

A. 1
B. 2
C. 3
D. None of these

Answer

To solve the problem, we need to analyze the sequence of complex numbers defined by:

1. \( z_1 = -1 - i \)
2. \( z_{n+1} = z_n^2 - i \) for all \( n \geq 1 \)

We are tasked with finding \( |z_{2013}| \).

### Step 1: Calculate \( z_2 \)

Using the formula for \( z_{n+1} \):

\[ z_2 = z_1^2 - i \]

Calculating \( z_1^2 \):

\[ z_1^2 = (-1 - i)^2 = (-1)^2 + 2(-1)(-i) + (-i)^2 = 1 + 2i - 1 = 2i \]

Now substituting back into the equation for \( z_2 \):

\[ z_2 = 2i - i = i \]

### Step 2: Calculate \( z_3 \)

Next, we calculate \( z_3 \):

\[ z_3 = z_2^2 - i \]

Calculating \( z_2^2 \):

\[ z_2^2 = i^2 = -1 \]

Now substituting back into the equation for \( z_3 \):

\[ z_3 = -1 - i \]

### Step 3: Calculate \( z_4 \)

Now we calculate \( z_4 \):

\[ z_4 = z_3^2 - i \]

Calculating \( z_3^2 \):

\[ z_3^2 = (-1 - i)^2 = 1 + 2i - 1 = 2i \]

Now substituting back into the equation for \( z_4 \):

\[ z_4 = 2i - i = i \]

### Step 4: Identify the pattern

From the calculations, we have:

- \( z_1 = -1 - i \)
- \( z_2 = i \)
- \( z_3 = -1 - i \)
- \( z_4 = i \)

We can see that:

- \( z_1 = -1 - i \) (odd index)
- \( z_2 = i \) (even index)
- \( z_3 = -1 - i \) (odd index)
- \( z_4 = i \) (even index)

Thus, we can conclude that:

- For odd \( n \): \( z_n = -1 - i \)
- For even \( n \): \( z_n = i \)

### Step 5: Determine \( z_{2013} \)

Since \( 2013 \) is odd, we have:

\[ z_{2013} = -1 - i \]

### Step 6: Calculate the modulus \( |z_{2013}| \)

Now, we need to find the modulus:

\[ |z_{2013}| = |-1 - i| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Final Answer

Thus, the modulus \( |z_{2013}| \) is:

\[ \boxed{\sqrt{2}} \]