14th December 2008

This is another good one about complex numbers. I am not trying to get a solution. I am sure many of you can.
I got this question from a user(Integrations) here. (I modified that one slightly!)

I want to stress the importance geometry can play in solving questions of Complex numbers. And why the importance of understanding Argand plane's equivalence with the XY plane of coordinate Geometry!

Find the locus of z such that

|z.k + 2i| > √2 for all real values of k

and that z is in the 4th quadrant.

Note: I have changed the language of the question to take away all the complexities.. so that you get a much simpler thought process :) I hope this hleps.

51 Answers

62
Lokesh Verma ·

well varun it says that k spans all reals.!

1
skygirl ·

concentric circles with radii between 1(k+1/k)2 n 2(k+1/k)2 units with center at (0.-2(k+1/k)).. ??

62
Lokesh Verma ·

Rephrased the question.. see if that makes it more clear?

9
Celestine preetham ·

i think its called an appolinious circle

1
maddybaddie ·

is z a semi-circular ring with inner radius 1(k+1/k)2 n outer radius 2(k+1/k)2? (since z is in 4th quadrant )

62
Lokesh Verma ·

no maddy it is not a circle..

see the thing given is that k could be any real number..

i dont know appolinious circle.. but the final solution has nothing much to with circles ;)

9
Celestine preetham ·

i think its called an appolinious circle

1
maddybaddie ·

i meant semi-circle...excluding d part in d 3rd quadrant ?

62
Lokesh Verma ·

Thsi si the locus for

1<|z| < 2

given in the 4th quadrant.

1
skygirl ·

concentric circles with radii between 1 n 2 units with center at (0.-2).. ??

1
skygirl ·

oh sry...
the two points will be 1(k+1/k) n √2(k+1/k)...

62
Lokesh Verma ·

but it is given for all real values of k!!

where is value of k that u will take!!

it is not

1<|z.k/(k+1)| < 2

1
maddybaddie ·

is d centre (0-2i)?

62
Lokesh Verma ·

no maddie...

see the problem is that k is not a fixed number.

Example:

for a fixed k, kz gives another complex number

for all real values of k, kz will give infinitely many points (which turn out to be ....)

62
Lokesh Verma ·

no..

Celestine.. i think all of u have got confused.

there are 2 things.. both have seperate answers

1) find all x such that x>k for a fixed k

2) find x such that x>k for all real values of k.

all of u are answering using k as in part 1!

i want k as part 2... read the question carefully!

1
maddybaddie ·

1
varun ·

No such z ??

1
maddybaddie ·

dis is wat i meant

62
Lokesh Verma ·

well maddie this is not the correct thing..

well tried..

see the problem is that the center of the circle changes with every k

so the center is dynamic.

Wait i will make this question much simpler!!!

Let me rechange the question.. bcos it seems that this has become too tough!

9
Celestine preetham ·

excluing portion btw

x2+(y+1)2= 1 and 1/2

Zk/K+1 lies btw x2+y+22 = 1 and 2

Zk/k+1 is point dividing 0 and Z in ratio k:1
k=-1 gives z as midpt of kz/z+1 which is impossible

so all other sol are possible were z vector cuts the circles its possible upto tangent at outer circle givin x=-y

so its lower half of 4 quad excluding wat i wrote above

62
Lokesh Verma ·

celestine but the question has changed in that time! :(

sorry to shuffle the queston around!

But i was having a feeling that no one is understanding the essense..

so i removed a few things to make it simpler.

and now it has a much more elegant proof and will explain much better (with simplicity) what i wanted to as well :)

Your Answer

Close [X]