Difference between revisions of "2002 AIME I Problems/Problem 12"
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Iterating <math>F</math> we get: | Iterating <math>F</math> we get: | ||
| − | + | <cmath> | |
\begin{align*} | \begin{align*} | ||
F(z) &= \frac{z+i}{z-i}\\ | F(z) &= \frac{z+i}{z-i}\\ | ||
| Line 12: | Line 12: | ||
F(F(F(z))) &= \frac{\frac{z+1}{z-1}i+i}{\frac{z+1}{z-1}i-i} = \frac{\frac{z+1}{z-1}+1}{\frac{z+1}{z-1}-1} = \frac{(z+1)+(z-1)}{(z+1)-(z-1)} = \frac{2z}{2} = z. | F(F(F(z))) &= \frac{\frac{z+1}{z-1}i+i}{\frac{z+1}{z-1}i-i} = \frac{\frac{z+1}{z-1}+1}{\frac{z+1}{z-1}-1} = \frac{(z+1)+(z-1)}{(z+1)-(z-1)} = \frac{2z}{2} = z. | ||
\end{align*} | \end{align*} | ||
| − | + | </cmath> | |
From this, it follows that <math>z_{k+3} = z_k</math>, for all <math>k</math>. Thus | From this, it follows that <math>z_{k+3} = z_k</math>, for all <math>k</math>. Thus | ||
Latest revision as of 21:39, 21 November 2018
Problem
Let 
for all complex numbers 
, and let 
for all positive integers 
. Given that 
and 
, where 
and 
are real numbers, find 
.
Solution
Iterating 
we get:
From this, it follows that 
, for all 
. Thus
Thus 
.
See also
| 2002 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
