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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    
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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
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