The polynomial P ( x ) = x 6 + ( x − r ) 6 + ( x − 2 r ) 6 − 3 r has 6 complex non real zeros for real r . Find the value of r if it is given that the real part of one of it's imaginary zero is 4 .
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After being told I was correct for my answer 4 I put 4 into P ( x ) and solved for the roots. There were SIX complex roots NOT "only two complex roots". Something is wrong. The question has to be changed and this solution is invalid. All SIX complex roots have the Real Component of 4. They are:
4 + 1 1 . 9 2 1 5 i , 4 − 1 1 . 9 2 1 5 i , 4 + 4 . 0 9 0 3 i , 4 − 4 . 0 9 0 3 i , 4 + 1 . 0 7 0 8 i , 4 − 1 . 0 7 0 8 i
If my analysis is wrong, please tell me the value of β in the solution above where α + ι β is one of the conjugates in the 'only two complex solutions'
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We observe that P ( r + x ) = P ( r − x ) for all x .That means if one zero is ( r + S ) the corresponding other would be ( r − S ) . We also know that complex zeros always occurs in pair. Now
Let α + ι β be a root of the equation P ( x ) = 0 .
⇒ α − ι β is the other root.
Now let α + ι β = r + x ⇒ r = α − s , x = s + ι β for a real s .
⇒ r − x = α − 2 s − ι β is also a root. Now since there are only two complex roots we can say that α − 2 s − ι β = α − ι β ⇒ s = 0 ⇒ α = r
And hence r = 4