thx bhaiyya and manipal what abt the other two parts?? Q2 is ambiguous i think
If 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..
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8 Answers
bhai 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/√2 and c=d =-1/√2 or the other way around
Q1 (d)
Q2 ???
Q3 (b)
This 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)2+(dk+b)2 = 2
b2k2+d2+d2k2+b2 = 2
a2+d2+c2+b2=2
so |x|=|y|=1 is the only possibility.
Q1: |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.
@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
Hmm, 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.