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#1 Posted 5:52pm 12-02-09 Complex number...Q.1 If |z|=1 then prove that complex number denoted by [sqrt](1+z)/(1-z)[/sqrt] always lie in either of the two fixed perpendicular lines.. Q.2 If z[ss]1[/ss]+z[ss]2[/ss]+z[ss]3[/ss]=0. then prove that the angle between any two pair of complex number is greater than equal to 2π/3
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#2 Posted 5:53pm 12-02-09 Re: Complex number...Q 1 was solved sometime back on the forum.. I think it was given by eureka.. i will have to check..
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#3 Posted 5:54pm 12-02-09 Re: Complex number...ok.. khoj lete hain..
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#4 Posted 5:56pm 12-02-09 Re: Complex number...http://targetiit.com/iit_jee_forum/posts/complex_geometry_295.html
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#5 Posted 7:19pm 12-02-09 Re: Complex number...Q 1. z lies on a circle for which the line joining (1,0) and (-1,0) are the ends of a diameter. That means arg(1+z/1-z) = ±π/2. Hence the argument of the square root of the number is ± π/4. Thus, the numbers lie on one of two perpendicular lines. Q 2. Needs to be rephrased I guess. It should be: there exists a pair such that the angle between them is greater than or equal to 2π/3. Please check |
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#6 Posted 7:39pm 12-02-09 Re: Complex number...this proof for the first part is awesome :) I wonder how much I have to learn to equal prophet :)
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#7 Posted 7:39pm 12-02-09 Re: Complex number...but then this was doable ;)
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#8 Posted 8:23pm 12-02-09 Re: Complex number...BHAIYA HOW TO APPROACH GRAPHICALLY.... |
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#9 Posted 8:25pm 12-02-09 Re: Complex number...Rotation as done by Prophet is a graphical method if u understand it well enuf :)
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