1992 AMC 12 Problem 15

Attempt Problem 15 of the 1992 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1992 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

15.

Let i=−1.i=\sqrt{-1}. For n≥1,n\ge1, define a sequence of complex numbers by z1=0,zn+1=zn2+i. \begin{aligned} z_1&=0,\\ z_{n+1}&=z_n^2+i. \end{aligned} In the complex plane, how far from the origin is z111?z_{111}?
11
2\sqrt2
3\sqrt3
110\sqrt{110}
255\sqrt{2^{55}}
Answer: B
Concepts:complex numberrecurrenceperiodic sequencemodulus
Small hint:
Compute the first few terms rather than trying to expand z111z_{111}
Big hint:
Look for a cycle beginning with z3z_3
Solution:
The first terms are z1=0,z2=i,z3=−1+i,z4=−i,z5=−1+i. \begin{aligned} z_1&=0,\quad z_2=i,\quad z_3=-1+i,\\ z_4&=-i,\quad z_5=-1+i. \end{aligned} Hence the sequence alternates between −1+i-1+i at odd indices and −i-i at even indices from z3z_3 onward. Therefore z111=−1+i,z_{111}=-1+i, whose distance from the origin is (−1)2+12=2.\sqrt{(-1)^2+1^2}=\sqrt2.
Thus the correct answer is B.
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.