Challenging Locus Questions (1 Viewer)

[CM_Tutor]

OK, so you have that angle CAD is arg (DA / CA), which is correct, and you have DA and CA. Substitute in DA and CA, do some algebra, and you should be able to get that angle CAD is arg z - arg (z + 1), as required.

Getting angle CBD is, in principle, similar, but you do need to know how to rewrite arg (-u) in terms of arg u, where u is any complex number.

Can you get out part (b) now?

 

[CM_Tutor]

What about CBD?

 

[CM_Tutor]

Originally posted by Grey Council
god damn you to hell

Sorry - I didn't mean to rain on your parade.

 

[Grey Council]

Question reposted, just for easy reference. ^_^

Suppose that z is a complex number satisfying |z| > 1 and 0 < arg z < pi / 2.

(a) Sketch on the Argand diagram the points A, B, C and D which represent, respectively, z, -1, 1 / z, and 1.

(b) Show that angles CAD and CBD are both arg z - arg (z + 1)

(c) Hence, show that ABCD is a cyclic quadrilateral, and that the centre of the circle through A, B, C and D is [(|z|^2 - 1) / (2 * Im(z))] * i

 

[Grey Council]

Originally posted by CM_Tutor
Getting angle CBD is, in principle, similar, but you do need to know how to rewrite arg (-u) in terms of arg u, where u is any complex number.
[/B]

arg BD = 0
arg of BC = arg(1/z + 1)

the heck, i'm stuck again. lol

 

[CM_Tutor]

For angle CAD, you needed CA and DA. So, for CBD you need CB and DB. Des this help?

 

[Grey Council]

Done it. A bit dodgy solution, though. hrm

now to try part c

*is scared already*

 

[CM_Tutor]

Originally posted by Grey Council
Done it. A bit dodgy solution, though. hrm

now to try part c

*is scared already*

Post it up - I'll let you know if it's 'dodgy' as you put it, and how it can be tightened up.

Don't be scared of the first part of (c) - it can be done in one line.

 

[Grey Council]

now that I've looked at the solution again, i'm not sure. I think I made up a rule. ^__^

Arg BD = Arg (1+1) = Arg 2 = 0
Arg CB = Arg (-1 - 1/z)
= arg((-z-1)/z)
= arg (-(z+1)/z)
so I thought maybe because its negative, you can flip the whole thing (ie take the reciprocal)

lol, never make up stuff in maths.

I made a mistake though, I thought this:
arg(z^-1) = -arg(z), right?
wtf, i think I know what to do
its:
arg2 - arg(-1-1/z), right?
so thats:
-arg (-(z+1)/z)
= arg( z/z+1) ______ ( as in, kinda like logs, ya know, to the power of negative one thingo)
which is the required angle.
LOL, am i king or what? my subconscious mind works in wierd ways, i reckon. lol

 

[Grey Council]

as for C, part i:
CD subtends equal angles at B and A (just proven, right?)
therefore ABCD is a cyclic quad.

and thats your two lines for you.

and for part ii), i seriously don't know where to begin. I can look at Spice Girl's solution, but I'd rather not. I think, if you don't really mind, I want to be led through the solution.

lol, there are no vector questions in the coroneos book, so thats prolly why i'm so damn weak at this type of thing. plus this isn't all that easy a question, I s'pose. hehe

 

[CM_Tutor]

Originally posted by Grey Council
now that I've looked at the solution again, i'm not sure. I think I made up a rule. ^__^

Arg BD = Arg (1+1) = Arg 2 = 0
Arg CB = Arg (-1 - 1/z)
= arg((-z-1)/z)
= arg (-(z+1)/z)
so I thought maybe because its negative, you can flip the whole thing (ie take the reciprocal)

lol, never make up stuff in maths.

I made a mistake though, I thought this:
arg(z^-1) = -arg(z), right?
wtf, i think I know what to do
its:
arg2 - arg(-1-1/z), right?
so thats:
-arg (-(z+1)/z)
= arg( z/z+1) ______ ( as in, kinda like logs, ya know, to the power of negative one thingo)
which is the required angle.
LOL, am i king or what? my subconscious mind works in wierd ways, i reckon. lol

As I said above, to find angle CBD you need the vectors CB and DB. Let's start with my solution, and then look at your comments ...

Recall that the vector LM, from L to M, is m - l, where the points L and M represent the complex numbers l and m, respecitvely.

So, DB (from D to B) is -1 - 1 = -2.
Thus arg DB = arg -2 = pi

Similarly, CB is -1 - 1 / z = -(z + 1) / z
Thus arg CB = arg [-(z+1) / z] = pi + arg[(z + 1) / z], using the result that arg (-u) = pi + arg u

Now, angle CBD = arg DB - arg CB = pi - (pi + arg[(z + 1) / z])
= pi - pi - arg[(z + 1) / z]
= -[arg(z + 1) - arg z], using arg(u / v) = arg u - arg v
= arg z - arg(z + 1), as required.

Comments ...

1. You have the wrong vector for DB, but the correct one for CB - this has led you to try to fudge the answer. You should have instead taken the fact that you couldn't get the answer as a warning to go back and look for an error - don't try to fudge in 4u, you markers will see through it.

2. You asked whether "-arg (-(z+1)/z) = arg( z/z+1) ______ ( as in, kinda like logs, ya know, to the power of negative one thingo) The result you were actually looking for was arg(-u) = pi + arg u. This is easily proven, as arg(-u) = arg(-1 * u) = arg(-1) + arg u = pi + arg u.

Note, however, that your observation of a connection between arg and log is an inciteful one. Several arg rules are similar to log rules, such as arg(u * v) = arg u + arg v and arg(u / v) = arg u - arg v, and arg(u^-1) = -arg u. This is not a coincidence - the two are related, but you'll have to do Uni maths to find out why.

Originally posted by Grey Council
as for C, part i:
CD subtends equal angles at B and A (just proven, right?)
therefore ABCD is a cyclic quad.

and thats your two lines for you.

and for part ii), i seriously don't know where to begin. I can look at Spice Girl's solution, but I'd rather not. I think, if you don't really mind, I want to be led through the solution.

lol, there are no vector questions in the coroneos book, so thats prolly why i'm so damn weak at this type of thing. plus this isn't all that easy a question, I s'pose. hehe

On the first part of (c), you are correct, but you could have written it in one line.

ie. CD subtends equal angles at B and A (proven in (b)), and so ABCD is a cyclic quad.

As for the second bit, forget complex numbers and think circles. Can you suggest any ways to find the centre of a circle which encloses a cyclic quad ABCD?

 

[Grey Council]

Originally posted by CM_Tutor
Recall that the vector LM, from L to M, is m - l, where the points L and M represent the complex numbers l and m, respecitvely.

So, DB (from D to B) is -1 - 1 = -2.
Thus arg DB = arg -2 = pi

d'oh

Similarly, CB is -1 - 1 / z = -(z + 1) / z
Thus arg CB = arg [-(z+1) / z] = pi + arg[(z + 1) / z], using the result that arg (-u) = pi + arg u

oh k, thats the result you were referring to. I didn't know that. hehe

Now, angle CBD = arg DB - arg CB = pi - (pi + arg[(z + 1) / z])
= pi - pi - arg[(z + 1) / z]
= -[arg(z + 1) - arg z], using arg(u / v) = arg u - arg v
= arg z - arg(z + 1), as required.

right. hehe

Comments ...
1. You have the wrong vector for DB, but the correct one for CB - this has led you to try to fudge the answer. You should have instead taken the fact that you couldn't get the answer as a warning to go back and look for an error - don't try to fudge in 4u, you markers will see through it.

lol, i was actually trying to guess the arg(-u) thing. hehe, but I see your point, my bad.

2. You asked whether "-arg (-(z+1)/z) = arg( z/z+1) ______ ( as in, kinda like logs, ya know, to the power of negative one thingo) The result you were actually looking for was arg(-u) = pi + arg u. This is easily proven, as arg(-u) = arg(-1 * u) = arg(-1) + arg u = pi + arg u.

ah, i see.

Note, however, that your observation of a connection between arg and log is an inciteful one. Several arg rules are similar to log rules, such as arg(u * v) = arg u + arg v and arg(u / v) = arg u - arg v, and arg(u^-1) = -arg u. This is not a coincidence - the two are related, but you'll have to do Uni maths to find out why.

Looking forward to it, although why the hell would angles be related to logs. hrm

On the first part of (c), you are correct, but you could have written it in one line.
ie. CD subtends equal angles at B and A (proven in (b)), and so ABCD is a cyclic quad.

bah to you. I COULD have written it in one line, but I tried to please you and wrote it in TWO. lol, that reminds me of Anton Chekhov's short story "Oh! The Public". If you haven't read it, I would ask you to read it. Its not long, and its humorous, so you'll enjoy it.
http://chekhov2.tripod.com/039.htm
but you'll see my point after you've read the story.

As for the second bit, forget complex numbers and think circles. Can you suggest any ways to find the centre of a circle which encloses a cyclic quad ABCD? [/B]

hrm, will spend one whole hour on the morning trip to school tommorow. hehe, i leave my house at 6:35ish, so i need to go to sleep early. hehe

 

[Grey Council]

ie. CD subtends equal angles at B and A (proven in (b)), and so ABCD is a cyclic quad.

As for the second bit, forget complex numbers and think circles. Can you suggest any ways to find the centre of a circle which encloses a cyclic quad ABCD?

well, i can see that 1 and -1 is a chord of the circle. and the y axis cuts straight through it at a 90 degree angle. so the centre lies on the y axis.

so the coordinates so far will be something like:
( 0 , a )

we know that the chord CD subtends equal angles at B and A, arg z - arg(z + 1) = arg ( z / (z+1) )
so if we can find the place where it is twice that, subtended from CD, that will be the centre. theoretically.
but i don't know how to start. and won't there be heaps of places subtended by CD that'll have an angle twice that of CAD? hrm

 

[CM_Tutor]

Originally posted by Grey Council
well, i can see that 1 and -1 is a chord of the circle. and the y axis cuts straight through it at a 90 degree angle. so the centre lies on the y axis.

so the coordinates so far will be something like:
( 0 , a )

Stick with this idea. Remember that we are talking about a circle. So, how far is it from (0, a) to A? to B? to C? to D?

we know that the chord CD subtends equal angles at B and A, arg z - arg(z + 1) = arg ( z / (z+1) )
so if we can find the place where it is twice that, subtended from CD, that will be the centre. theoretically.
but i don't know how to start. and won't there be heaps of places subtended by CD that'll have an angle twice that of CAD? hrm

Yes, there will - they will be all the points on a circle throgh C, D and the point (0, a) - but that doesn't necessarily matter, as you know the one you are looking for is on the imaginary axis. However, I'm not sure how you'd proceed from here - stick to the circle approach above.

 

[Grey Council]

Originally posted by CM_Tutor
Stick with this idea. Remember that we are talking about a circle. So, how far is it from (0, a) to A? to B? to C? to D?

radii, all of them. hrm

 

[CM_Tutor]

So, taking the point you're trying to find, (0, a), as P, then PA = PB = PC = PD (equal radii). Two of these, PB and PD, are easily expressed in terms of a. We need a linked to z, so equate one of these to PA. You may need to put
z = x + iy to sort it all out...

 

[Grey Council]

got it out:
equate distances between C to 1 and C to z
sqrt ( 1 + a^2 ) = sqrt (x^2 + (y-a)^2)
1 = x^2 + y^2 - 2ya
juggle equation:
c = (x^2 + y^2 - 1)/2y
well, since I forgot the i (just used normal coordinate axis), we know c lies in y axis, so you can just add an i.

yay, now I can try some of your other questions, CM_Tutor. ^_^

 

[CM_Tutor]

Well done.

 

[CM_Tutor]

4. CHALLENGE!
Prove that (|z| - iz)/(|z| + iz) = -i(sec @ + tan @), where r(z) =/= 0 and arg z = @.

A message I received prompted me to have another look at this question, and I came up with a geometric approach that is similar to Oldman's, but not identical, so I thought I'd share it.

Start by drawing a circle, centre at O - the origin - on the Argand Diagram, with radius |z|. Label the point where
z = |z| as A - this is where the circle meets the positive real axis, and the point iz (on the circle) as C. Construct the parallelogram OABC - B represents the complex number |z| + iz, and is not on the circle.

It is clear that the vector OB is |z| + iz, and the vector CA is |z| - iz, and thus the required expression is
vector CA / vector OB.

Now, OA = OC (equal radii), and so the parallelogram OABC is, in fact, a rhombus. Thus, the diagonals are perpendicular, and hence vector CA / vector OB = -i * CA / OB.

Angle COA = arg(iz) = arg i + arg z = 90 + @, and so angle OCB = 90 - @ (Co-interior angles on parallel lines). Since the diagonals bisect the angles, angle OCA = 45 - @ / 2 and angle OBA = 45 + @ / 2.

Applying the sine rule in triangle OBA, OA = OB * sin(45 + @ / 2) / sin(90 - @), and applying the sine rule in
triangle OCA, OA = CA * sin(45 - @ / 2) / sin(90 + @).

Equating these expressions for OA, and rearranging, we get
CA / OB = sin(90 + @) * sin(45 + @ / 2) / [sin(90 - @) * sin(45 - @ / 2)]
= cos@ * (1 / sqrt(2)) * [cos(@ / 2) + sin(@ / 2)] / {cos@ * (1 / sqrt(2)) * [cos(@ / 2) - sin(@ / 2)]}
= [cos(@ / 2) + sin(@ / 2)]² / {[cos(@ / 2) - sin(@ / 2)] * [cos(@ / 2) + sin(@ / 2)]}
= [cos²(@ / 2) + sin²(@ / 2) + 2cos(@ / 2)sin(@ / 2)] / [cos²(@ / 2) - sin²(@ / 2)]
= [1 + sin(2@ / 2)] / cos(2@ / 2)
= sec@ + tan @

So, (|z| - iz) / (|z| + iz) = -i(sec@ + tan@), as required.