Difference between revisions of "2020 AMC 12B Problems/Problem 23"
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==Solution== | ==Solution== | ||
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| + | For <math>n=2</math>, we see that if <math>z_{1}+z_{2}=0</math>, then <math>z_{1}=-z_{2}</math>, so they are evenly spaced along the unit circle. | ||
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| + | For <math>n=3</math>, WLOG, we can set <math>z_{1}=1</math>. Notice that now <math>\mathscr{Re}(z_{1}+z{2})=-1</math> and <math>\mathscr{Im}\{z_{1}\}=-\mathscr{IM}\{z_{2}\}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 00:57, 8 February 2020
Problem 23
How many integers 
are there such that whenever 
are complex numbers such that
![\[|z_1| = |z_2| = ... = |z_n| = 1 \text{ and } z_1 + z_2 + ... + z_n = 0,\]](http://latex.artofproblemsolving.com/3/b/5/3b5ec1e5f9204ea20f9438cf2c48b0c9464f27ac.png)
then the numbers are equally spaced on the unit circle in the complex plane?
Solution
For 
, we see that if 
, then 
, so they are evenly spaced along the unit circle.
For 
, WLOG, we can set 
. Notice that now $\mathscr{Re}(z_{1}+z{2})=-1$ (Error compiling LaTeX. Unknown error_msg) and $\mathscr{Im}\{z_{1}\}=-\mathscr{IM}\{z_{2}\}$ (Error compiling LaTeX. Unknown error_msg)
See Also
| 2020 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 22 |
Followed by Problem 24 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
