# a little bit hard complex number questionWatch

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#1
in this question, all the numbers can be complex.
0
1 month ago
#2
Have you got anywhere?
How is O expressed in terms of the 2nd quadratic?

(Original post by cxs)
in this question, all the numbers can be complex.
0
#3

that is what i got. And it lies on the same circle means that if we can prove the "equally inclined", we can derive the last part.
But I have no idea how to prove that it is equally inclined.
(Original post by mqb2766)
Have you got anywhere?
How is O expressed in terms of the 2nd quadratic?
Last edited by cxs; 1 month ago
0
1 month ago
#4
Yes, that looks good. I'll have a look at it a bit later, busy this morning.
Is this from Hardy's book/tripos?
(Original post by cxs)

that is what i got. And it lies on the same circle means that if we can prove the "equally inclined", we can derive the last part.
But I have no idea how to prove that it is equally inclined.
0
#5
Thanks.
Yeah, it is in the CH3
0
1 month ago
#6
From a quick read of Hardy, he pretty much has the solution preceding the question? Not been carefully through it, but the transformation where he maps
(z1 + z2)(z3 + z4) = 2(z1z2 + z3z4)
to
(z1 - 1/2(z3 + z4))(z2 - 1/2(z3 + z4)) = (1/2 (z3 - z4))^2
(Original post by cxs)
Thanks.
Yeah, it is in the CH3
0
#7
OMG I think you're absolutely right!!
I did not notice that.....
THanks !
(Original post by mqb2766)
From a quick read of Hardy, he pretty much has the solution preceding the question? Not been carefully through it, but the transformation where he maps
(z1 + z2)(z3 + z4) = 2(z1z2 + z3z4)
to
(z1 - 1/2(z3 + z4))(z2 - 1/2(z3 + z4)) = (1/2 (z3 - z4))^2
0
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