IIT JEE , AIEEE Forum - Mathematics , Trignometry

m.viddya

Q.z_1,z₂,z₃ are in harmonic progression then z_1,z_2,z₃ lie on

1.a straight line

2.an ellipse

3.a parabola

4.none.

PHYCHEMATICO

correct answer is A. STRAIGHT LINE.

m.viddya

can i have the solution plzzz...

hsbhatt

I think they lie on a circle.

We have 1/z₂ = 1/2(1/z₁+1/z₃)
Simplifying, we get

z₁z₂+z₂z₃ = 2z₁z₃

or (z₂-z₃)/(z₁-z₂) = z₃/z₁.

Thus, arg(z₂-z₃/z₁-z₂) = arg(z₃/z₁).

Lets plot the points geometrically.

Say z₁ = A, z₂ = B and z₃ = C.

Now using the fact that 1/z₂ = 1/2(1/z₁+1/z₃) you can prove that

1.either |z₁|>=|z₂|>=|z₃| or |z₁|<=|z₂|<=|z₃|.Using |1/z₂| = 1/2|1/z₁+1/z₃|. (Try to prove it).
2. z₃ never lies between z₁ and z₂. Using argz₂ = arg(z₁)+arg(z₃) - arg(z₁+z₃).
(Highlighted portion is wrong)
1. z₂ always lies between z₁ and z₃. Using argz₂ = arg(z₁)+arg(z₃) - arg(z₁+z₃).

2. arg(z₂-z₃/z₁-z₂) = arg(z₃/z₁) excludes cases when OABC is not convex (both have the same sense, clockwise or anticlockwise. This tells us CB cannot 'tilt behind' AB.

Hence, OABC forms a convex quadrilateral.

And from arg(z₂-z₃/z₁-z₂) = arg(z₃/z₁) we get that angle COB and angle CBA are supplementary. Hence, OABC is a cyclic quadrilateral

That is A,B, and C lie on a circle passing through origin.

Proving that OABC is a convex quadrilateral is a critical part of the proof.

m.viddya

well the answer is given as ellipse

feynmann

AP = k BP

consider A, B to be fixed

P describes a circle .

So a circle .

Greatdreams

I also feel that its a circle.

Greatdreams

Consider the angles at the points z1 and z3 to be some x and y

Now you have by the euler's form e ix = -z1/z2 - z1 and e iy = z2 - z3/-z3

now multiply them and you have z1z2 - z1z3 = (z2z3 - z1z3)e i(x+y)

If the points lie on a circle then the quadrilateral formed

should be cyclic that is opposite angles should sum up to

180 degrees.

Then you have x + y = 180 degrees

So z1z2 - z1z3 = (z2z3 - z1z3)e i

= (z2z3 - z1z3)*-1

Or z1z2 + z2z3 = 2z1z3

Divide both sides by z1z2z3 and you get the required

condition.

You thus get 1/z3 + 1/z1 = 2/z2

So z1,z2,z3 are in H.P..........

hsbhatt

Feynmann :I am a little rusty with my fundas but does the circle you describe pass thru A and B. I think not

feynmann

Yes u are right ( I was in a li'l bit of hurry ), but still my answer remains the same !!

Here we have

(z2-z1)/(z2 - z3 ) = -z1/z3

so we get

arg(z2-z1)/(z2 - z3 ) = arg(-z1/z3)

Now as mentioned in my previous post consider z1 & z3 to be fixed

so it means that argument of LHS is constant .

which in turn means that z2 lies on a portion of the circle of which z1z3 is a chord .

It may be a st. line only in a special case .

animal

hey man i think this problem can be done by taking z₁=cos $\alpha$ +isin $\alpha$ and similarly z2 and z3.

tell me if i m correct..

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