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

Home > Just Math
Angle Ascertainment (Posted on 2015-01-03) Difficulty: 3 of 5
z is a complex number and P is the product of the roots of the equation:
z6 + z4 + z3 + z2 + z2 +1 = 0 with each root having a positive imaginary part.

Given that, P = r(cos θ + i*sin θ), where r > 0 and 0 ≤ θ < 360o

Find θ.

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Hints/Tips re: the roots of the imaginary att K.Sengupta Comment 2 of 2 |
(In reply to the roots of the imaginary by Dej Mar)

Hi, Dej Mar

IMHO, It was not z6 + z4 + z3 + 2z2 +1 = 0  to be solved:

it was either z6 + z4 + z3 + z+ z + 1 = 0        (i)

or z6 +z5 + z4 + z3 + z+ z+ 1 = 0                 (ii)

(ii) is simple and interesting

(i) more difficult and less interesting

Both have positive and negative imaginary  parts, in (ii) you can get the product  and the angle without listing the roots.
So - either solve both corrected version or wait for KS's reply. 

Still - may be I'm wrong, but it was a funny way to write 2z2....


  Posted by Ady TZIDON on 2015-01-04 02:28:51
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 (4)
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