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#2 2013-01-06 20:20:11
- scientia
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Re: Complex no.
Let the roots be , , , so . Then:
So (i) and (iii) are true.
We have so is true. The others are not necessarily true: e.g. has complex roots and .
Last edited by scientia (2013-01-18 22:43:43)
#3 2013-01-07 17:35:53
- noelevans
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Re: Complex no.
(B) When a=c we obtain ax²+bx+a= 0 which has roots
-b±√(b²-4a²)
2a
And multiplying these two roots together gives b²-(b²-4a²) = 4a²/4a² = 1.
4a²
So the two roots are reciprocals of each other, BUT the Discriminant is not negative unless
b²<4a². So |b|<|2a| is required for the roots to be complex (not real roots).
So the first condition would be true if it were |b|<|2a| instead of |b|<|a|. 
From this problem we see that there are infinitely many complex number pairs that are complex
conjugates AND reciprocals at the same time. But there is only ONE pair of complex numbers that
are both OPPOSITES AND RECIPROCALS at the same time. And that would be ...
Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).
LaTex is like painting on many strips of paper and then stacking them to see what picture they make.