a little bit hard complex number question Watch

cxs
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in this question, all the numbers can be complex.
\begin{array}{l}{\text {  If the points } A_{1}, A_{2} \text { are given by } a z^{2}+2 b z+c=0, \text { and the points } A_{3}, A_{4}} \\ {\text { by } a^{\prime} z^{2}+2 b^{\prime} z+c^{\prime}=0, \text { and } O \text { is the middle point of } A_{3} A_{4}, \text { and } a c^{\prime}+a^{\prime} c-2 b b^{\prime}=0,} \\ {\text { then } O A_{1}, O A_{2} \text { are equally inclined to } A_{3} A_{4} \text { and } O A_{1} \cdot O A_{2}=O A_{3}^{2}=O A_{4}^{2}}\end{array}
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mqb2766
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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.
\begin{array}{l}{\text {  If the points } A_{1}, A_{2} \text { are given by } a z^{2}+2 b z+c=0, \text { and the points } A_{3}, A_{4}} \\ {\text { by } a^{\prime} z^{2}+2 b^{\prime} z+c^{\prime}=0, \text { and } O \text { is the middle point of } A_{3} A_{4}, \text { and } a c^{\prime}+a^{\prime} c-2 b b^{\prime}=0,} \\ {\text { then } O A_{1}, O A_{2} \text { are equally inclined to } A_{3} A_{4} \text { and } O A_{1} \cdot O A_{2}=O A_{3}^{2}=O A_{4}^{2}}\end{array}
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cxs
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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?
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mqb2766
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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)
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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.
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cxs
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Thanks.
Yeah, it is in the CH3
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mqb2766
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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
looks key for the part you're asking about?
(Original post by cxs)
Thanks.
Yeah, it is in the CH3
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cxs
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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
looks key for the part you're asking about?
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