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#1 Posted 8:13pm 01-03-09 paragraphIf X and Y are 2 complex numbers such that lXl ≤1 and lYl ≤1 and lX+iYl=lX-iYl=2. Q1. Which of the following is true about lXl and lYl ? (a) lXl=lYl=1/2 (b) lXl=1/2 and lYl=3/4 (c) lXl=lYl=3/4 (d) lXl=lYl=1 Q2. Which of the following is true for X and Y? (a) Re(X) = Re(Y) (b) Im(X)=Im(Y) (c) Re(X)=Im(Y) (d) Im(X)=Re(Y) Q3. Number of complex numbers X satisfying the above conditions is (a) 1 (b) 2 (c) 4 (d) Indeterminate Again i dont have solns reason is the same as my trig. question..
IIT has walked to me ... off to IIT K |
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#2 Posted 8:58pm 01-03-09 Re: paragraphbhai dekh put X=a+ib Y=c+id and solve u will get a/b=c/d and usinf AM≥GM and we get ab≤1/2 and similarly cd≤1/2 so we get a=b=1/[sqrt]2[/sqrt] and c=d =-1/[sqrt]2[/sqrt] or the other way around Q1 (d) Q2 ??? Q3 (b)
There are two things you should never try to prove ...... the impossible and the obvious ;) |
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#3 Posted 9:22pm 01-03-09 Re: paragraphThis is a very good question... x=a+ic y=b+id x+iy=a-d+i(c+b) x-iy=a+d+i(c-b) modulus is same.. hence 2bc=2ad or a/b=c/d = k let a= bk, c=dk also, modulus is 2 so (bk-d)[p]2[/p]+(dk+b)[p]2[/p] = 2 b[p]2[/p]k[p]2[/p]+d[p]2[/p]+d[p]2[/p]k[p]2[/p]+b[p]2[/p] = 2 a[p]2[/p]+d[p]2[/p]+c[p]2[/p]+b[p]2[/p]=2 so |x|=|y|=1 is the only possibility.
-when spring comes, it melts the snow one flake at a time… |
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#4 Posted 9:28pm 01-03-09 Re: paragraphthx bhaiyya and manipal what abt the other two parts?? Q2 is ambiguous i think
IIT has walked to me ... off to IIT K |
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#5 Posted 2:22pm 02-03-09 Re: paragraphQ1: |x+iy| ≤ |x|+|iy| = |x|+|y| ≤ 1+1 = 2. |x-iy| ≤ |x|+|iy| = |x|+|y| ≤ 2. So |x+iy| = |x-iy| = 2 implies |x| = |y| = 1. That means x and y lie on a circle of radius 1 centred at (0,0) x and iy must lie on oppoiste ends of a diameter of this circle i.e. iy = -x From this we infer that y = ix. It follows that Im(y) = Re(x) [Option c] It is also easy to see that since all we need is that |x| = |y| = 1 and y=ix, there are infinitely many points. |
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#6 Posted 2:32pm 02-03-09 Re: paragraph@prophet... do u know any site that can teach how to approach complex no.s graphically??it will be really helpful for all of us |
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#7 Posted 2:44pm 02-03-09 Re: paragraphHmm, I've not seen any site as such. One good resource is a book named "Complex Numbers from A to Z". The author is Titu Andreescu. Pirated versions of the book can be downloaded free from the net. This is a very good read because he starts from fundamentals and then takes you very deep into the topic. The author, by the way, trained the USA IMO "dreamteam" in which every team member got full 42 points! So rest assured this is a quality book. |
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#8 Posted 2:46pm 02-03-09 Re: paragraphhi bhaiyya.........
"Make ur Mind Simple.......dont loss it on the Path of ur Life"..... |
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#9 Posted 2:52pm 02-03-09 Re: paragraphthank you very much! |
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