All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Complex Number and Power Puzzle (Posted on 2015-11-23) Difficulty: 3 of 5
Determine all values of θ with 0 ≤ θ ≤ 360o for which the complex number
z = cos θ + i*sin θ satisfies z28 - z8 -1 = 0

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution Comment 1 of 1
If you use the identity z^n = cos(nθ)+isin(nθ) the equation becomes
cos(28θ)+isin(28θ)-cos(8θ)-isin(8θ)-1=0

Separating this into real and imaginary components gives the system:
cos(28θ)-cos(8θ)-1=0
sin(28θ)-sin(8θ)=0

Note that each of these is periodic with period 360/4 = 90 degrees (4 being the GCD of 8 and 28.)
This could be used along with a lot of identities to reduce and solve analytically, but I didn't feel like it.  So I just made a table to get the solutions of each on the interval [0,90)
The cosines equation has 8 solutions and the sine has 12.  The two in common are 15 and 75.

Add the period to give all the solutions sought:
θ={15,75,105,165,195,255,285,345}

  Posted by Jer on 2015-11-26 09:51:41
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (9)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information