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: | ||
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\begin{align*} | \begin{align*} | ||
F(z) &= \frac{z+i}{z-i}\\ | F(z) &= \frac{z+i}{z-i}\\ | ||
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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*} | ||
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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 | ||
Revision as of 21:30, 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:
\begin{align*}
F(z) &= \frac{z+i}{z-i}\\
F(F(z)) &= \frac{\frac{z+i}{z-i}+i}{\frac{z+i}{z-i}-i} = \frac{(z+i)+i(z-i)}{(z+i)-i(z-i)}= \frac{z+i+zi+1}{z+i-zi-1}= \frac{(z+1)(i+1)}{(z-1)(1-i)}\\
&= \frac{(z+1)(i+1)^2}{(z-1)(1^2+1^2)}= \frac{(z+1)(2i)}{(z-1)(2)}= \frac{z+1}{z-1}i\\
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*}
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. 
