Question

Let i = $\sqrt{-1}$ Define a sequence if complex by $z_1$ = 0, $z_{n+1}$ = $z_n^2$ + i for n$\geq$1 Then which of the following are true.

• A. |$z_{2050}$| = $\sqrt{3}$
• B. |$z_{2017}$| = $\sqrt{2}$
• C. |$z_{2016}$| = 1
• D. |$z_{2111}$| = $\sqrt{2}$

Answer 1

Answer: (B), (C), (D)

|$z_{2017}| = \sqrt{2}$, |$z_{2016}| = 1$, and |$z_{2111}| = \sqrt{2}$

Answer 2

The sequence of complex numbers is defined by $z_1 = 0$ and $z_{n+1} = z_n^2 + i$ for $n \geq 1$. We compute the first few terms of the sequence:

$z_1 = 0$ $z_2 = z_1^2 + i = 0^2 + i = i$ $z_3 = z_2^2 + i = i^2 + i = -1 + i$ $z_4 = z_3^2 + i = (-1+i)^2 + i = (1 - 2i + i^2) + i = (1 - 2i - 1) + i = -2i + i = -i$ $z_5 = z_4^2 + i = (-i)^2 + i = i^2 + i = -1 + i$ $z_6 = z_5^2 + i = (-1+i)^2 + i = -i$

We observe a pattern for $n \geq 3$: the sequence alternates between $-1+i$ and $-i$.

Specifically, for $n \geq 3$: If $n$ is odd, $z_n = -1+i$. If $n$ is even, $z_n = -i$.

Now, let's compute the magnitudes of these terms: $|z_1| = |0| = 0$ $|z_2| = |i| = 1$ $|z_3| = |-1+i| = \sqrt{(-1)^2 + 1^2} = \sqrt{1+1} = \sqrt{2}$ $|z_4| = |-i| = 1$ $|z_5| = |-1+i| = \sqrt{2}$ $|z_6| = |-i| = 1$

The pattern for the magnitudes for $n \geq 3$ is: If $n$ is odd, $|z_n| = \sqrt{2}$. If $n$ is even, $|z_n| = 1$.

Now we check the given statements:

1. |$z_{2050}| = \sqrt{3}$: Here $n=2050$. Since $2050 \geq 3$ and 2050 is even, |$z_{2050}| = 1$. The statement |$z_{2050}| = \sqrt{3}$ is false.
2. |$z_{2017}| = \sqrt{2}$: Here $n=2017$. Since $2017 \geq 3$ and 2017 is odd, |$z_{2017}| = \sqrt{2}$. The statement |$z_{2017}| = \sqrt{2}$ is true.
3. |$z_{2016}| = 1$: Here $n=2016$. Since $2016 \geq 3$ and 2016 is even, |$z_{2016}| = 1$. The statement |$z_{2016}| = 1$ is true.
4. |$z_{2111}| = \sqrt{2}$: Here $n=2111$. Since $2111 \geq 3$ and 2111 is odd, |$z_{2111}| = \sqrt{2}$. The statement |$z_{2111}| = \sqrt{2}$ is true.

The true statements are |$z_{2017}| = \sqrt{2}$, |$z_{2016}| = 1$, and |$z_{2111}| = \sqrt{2}$.