complex numbers.... need help with a question? (1 Viewer)

[SXC_DRAGON_BOY]

3 complex numbers Z1, Z2 and Z3 lie on a circle passing through the origin, show that 1/Z1, 1/Z2 and 1/Z3 are collinear

 

[Rahul]

jus lookin at it:

ummm...

if z1, z2, z3 lie on a circle then their modulus must be the same

therefore,
z1=a+bi
z2=a-bi
z3=-a-bi

hmmmm......i have 4gotten to do this:mad1: i guess i gotta go back and study

i think i am on the right track.....i think!

 

[SXC_DRAGON_BOY]

thanks for trying

this is a funny question cause my teacher told me that a selective skool in my region couldnt do this one...
i know it has to do with the modulus, but just what?

 

[ND]

Originally posted by Rahul

if z1, z2, z3 lie on a circle then their modulus must be the same

That would be true if the centre of the circle was the origin, however, it says the circle passes through the origin. I don't have any ideas. I don't see how its do-able considering that the locus for any Z is lZl <= R, -pi <= Arg(Z) <= pi. Are you sure the question is correct?

 

[phenol]

well if the centre of the circle is not the origin then we can do x-y plane shift

define

y' = y + h
x' = x + k

assume that (k, h) is the centre of the circle

then we now have a new x'-y' plane where the circle has centre on origin.

then

let z1, z2, z3 be rcis(pi), rcis(xi), rcis(nu) respectively then

1/z1 = cis(-pi) / r^2
similarly
1/z2 = cis(-xi)/ r^2
1/z3 = cis(-nu)/ r^2

easily see the three points are NOT colinear rather they lie on the same circle with radius 1/(r^2) and origin (k, h)

pi

 

[:: ck ::]

i c sun yan zi

 

[RIZAL]

*thinks*

I'll put up the answer tonight.

 

[RIZAL]

hmmmm, too hard - I'll get my old man to have a crack at it.

 

[OLDMAN]

A hard question. Use this to review circle geometry and complex vectors and their args.

Three complex numbers p,q and r are collinear in the Argand plane if arg(p-q)-arg(q-r)=0,pi,-pi (0,180,-180). This just means that the two vectors p-q and q-r have the same arg or are supplementary.

Three points 1/z1,1/z2, and 1/z3 are collinear if
arg(1/z1-1/z2)-arg(1/z2-1/z3)=0,pi, or -pi
=arg((z2-z1)/z1z2)-arg((z3-z2)/z2z3)
= arg(z2-z1)-arg(z1z2)-arg(z3-z2)+arg(z2z3)
=arg(z2-z1)-arg(z3-z2)-arg(z1)-arg(z2)+arg(z2)+arg(z3)
=(arg(z2-z1)-arg(z3-z2))-((arg(z1)-arg(z3))

But arg(z2-z1)-arg(z3-z2) is either the -angle z1z2z3 or the external angle at z2 of the cyclic quad. Oz1z2z3 depending on how you draw these concyclic points.

arg(z1)-arg(z3) is an interior angle z1Oz2 of cyclic quad. Oz1z2z3.
The expression (arg(z2-z1)-arg(z3-z2))-((arg(z1)-arg(z3)) will give you the negative sum of two interior opposite angles -pi, or the difference of an external angle and an opposite interior angle, zero. Proven.

 

[spice girl]

Idunno if my solution is the same as OLDMAN's, but anyway:

When comparing z to 1/z, what's not important here is arg(1/z) = -arg(z)

What's more important is that |1/z| = 1/|z|

So If z1, z2, z3 represented by A, B, C, (origin O as usual), we have cyclic quad OABC (without loss of generality assume OABC is in that order).

Then suppose 1/z1(bar), 1/z2(bar), 1/z3(bar) is represented by A', B', C'.

Note that OA'A , OB'B, OC'C are collinear since arg(1/z1(bar)) = -arg(1/z1) = arg(z1)

Consider triangles OAB and OA'B':
angle O is common
OA/OB' = OB/OA' = ab

Thus OAB similar to OA'B'
and angle OAB = angle OB'A' (corresponding angles of similar triangles equal)

similarly, OBC is similar to OB'C' and angle OB'C' = OCB

(If you aren't following, it's prolly cos u aren't drawing this)

But angle OAB + angle OCB = pi (opp angles of cyclic quad supplementary)

Thus A', B', C' are collinear (angle OB'A' + angle OB'C' = pi)

Now, since 1/z1 is 1/z1(bar) reflected about the x axis, etc, thus since A', B', C' are collinear, thus 1/z1, 1/z2, 1/z3 are collinear (the line a reflection of A'B'C' about the x axis)

 

[OLDMAN]

Well done Spice Girl. Shows what a good grasp of the basic algebra and geometry of complex reciprocals and conjugates can achieve.