Let `i=sqrt(-1)` Define a sequence of complex number by `z_1=0, z_(n+1) = (z_n)^2 + i` for `n>=1`. In the complex plane, how far from the origin is `z_111`?

Options

A. `1`
B. `2`
C. `3`
D. `4`

Explanation

To solve the problem, we will define the sequence of complex numbers step by step and find the distance of \( z_{111} \) from the origin.

### Step 1: Define the sequence
We start with the initial value:
\[ z_1 = 0 \]
The recursive formula for the sequence is given by:
\[ z_{n+1} = z_n^2 + i \]

### Step 2: Calculate the first few terms
Let's calculate the first few terms of the sequence:

- For \( n = 1 \):
\[ z_2 = z_1^2 + i = 0^2 + i = i \]

- For \( n = 2 \):
\[ z_3 = z_2^2 + i = i^2 + i = -1 + i \]

- For \( n = 3 \):
\[ z_4 = z_3^2 + i = (-1 + i)^2 + i \]
Expanding \( (-1 + i)^2 \):
\[ (-1)^2 + 2(-1)(i) + i^2 = 1 - 2i - 1 = -2i \]
Therefore,
\[ z_4 = -2i + i = -i \]

- For \( n = 4 \):
\[ z_5 = z_4^2 + i = (-i)^2 + i = -1 + i \]

- For \( n = 5 \):
\[ z_6 = z_5^2 + i = (-1 + i)^2 + i = -2i + i = -i \]

### Step 3: Identify the pattern
From the calculations, we observe:
- \( z_3 = -1 + i \)
- \( z_4 = -i \)
- \( z_5 = -1 + i \)
- \( z_6 = -i \)

This indicates a repeating pattern:
- \( z_{2n-1} = -1 + i \) for odd \( n \)
- \( z_{2n} = -i \) for even \( n \)

### Step 4: Determine \( z_{111} \)
Since \( 111 \) is odd, we can use the pattern:
\[ z_{111} = z_3 = -1 + i \]

### Step 5: Calculate the distance from the origin
The distance from the origin to the complex number \( z_{111} \) is given by the modulus:
\[ |z_{111}| = |-1 + i| = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Final Answer
The distance from the origin to \( z_{111} \) is:
\[ \sqrt{2} \]