Challenging Locus Questions (1 Viewer)

[Grey Council]

1. If z = x + iy, express Z = (z−1)/z in the form X + iY (where X, Y are real). Hence or otherwise, prove that if the point representing z on the Argand diagram describes a unit circle about the origin, then the point representing Z also describes a unit circle, and find the centre of this circle.

2. If the argument of the complex number (z−1)/(z+1) is pi/4 , show that z lies on a fixed circle whose centre is at the point representing i.

3. If the point z moves on a semicircle, centre the origin and radius 2, in an anticlockwise direction from the point 2 to the point −2, find the path traced by
the point 1/z.

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

Hanyway, If you can do the last one, you are a genius. If you aren't a genius, don't even attempt it.

My question, apart from the solutions to the above questions, is should I bother doing this exercise? Or is it a bit too high for HSC standards?

 

[Grey Council]

*plaintively*
help!

 

[McLake]

*Brain explosion*

Let Keypad, turtle, OLDMAN or spice girl answer it ...

 

[CM_Tutor]

Originally posted by Grey Council
1. If z = x + iy, express Z = (z−1)/z in the form X + iY (where X, Y are real). Hence or otherwise, prove that if the point representing z on the Argand diagram describes a unit circle about the origin, then the point representing Z also describes a unit circle, and find the centre of this circle.

...

My question, apart from the solutions to the above questions, is should I bother doing this exercise? Or is it a bit too high for HSC standards?

This is a perfectly reasonable HSC question, albeit not an easy one. You can do it by an algebra bash, but that can be simplified substantially if you think a little first.

Ok, lets start two observations:

1. (z - 1) / z = (z / z) - (1 / z) = 1 - 1/z, which may be a more useful form.

2. 1 / z = z(bar) / |z|^2 ... If you don't know this, learn it and know how to prove it in two lines. It's a very useful result.

OK, so Z = (z - 1) / z = 1 - z(bar) / |z|^2 = 1 - (x - iy) / (x^2 + y^2) = X + iY
where X = 1 - x / (x^2 + y^2) and Y = y / (x^2 + y^2)

Now, the locus of z is the unit circle about O, and hence |z| = 1, and so x^2 + y^2 = 1.

Thus Z = X + iY where X = 1 - x, and Y = y.

All that remains is to prove that the locus of Z = 1 - x + iy = 1 - z(bar) is a unit circle, and find its centre - this is easiest to see geometrically.

Now, if z lies on the unit circle about 0, then so do z(bar) and -z(bar) - draw a diagram if you aren't convinced. 1 - z(bar) will therefore lie on the unit circle about 1, as adding 1 will simply shift the locus 1 unit along the real axis.

Alternately, if you prefer algebra, rewrite the equation as z(bar) = 1 - Z. We know |z| = 1, and hence |z(bar)| = 1. So, we have |1 - Z| = 1 = |Z - 1|. ie, a unit circle about 1.

 

[CM_Tutor]

Originally posted by Grey Council
3. If the point z moves on a semicircle, centre the origin and radius 2, in an anticlockwise direction from the point 2 to the point −2, find the path traced by
the point 1/z.

...

My question, apart from the solutions to the above questions, is should I bother doing this exercise? Or is it a bit too high for HSC standards?

This one is easy PROVIDED you correctly use the result I mentioned above, ie that

1 / z = z(bar) / |z|^2

Here, you know what |z| is. Now, think GEOMETRICALLY about the relationship between z and z(bar).

I'm going to let you think a bit more about this one...

 

[CM_Tutor]

Originally posted by Grey Council
2. If the argument of the complex number (z−1)/(z+1) is pi/4 , show that z lies on a fixed circle whose centre is at the point representing i.

My question, apart from the solutions to the above questions, is should I bother doing this exercise? Or is it a bit too high for HSC standards?

This is a common trial question, but usually asked in a slightly different way, with the angle ususally being pi / 2 - in which case the locus is a semicircle.

This is a problem in GEOMETRY, it is NOT a problem in algebra ...

Again, lets start with a couple of observations...

First, recall that arg(z / w) = arg z - arg w

Second, recall that z - z_0 is a vector from z_0 to z

Now, we have arg[(z - 1) / (z + 1)] = pi / 4
ie arg(z - 1) - arg(z + 1) = pi / 4

arg(z - 1) is the direction of the vector from 1 to z.

Now, draw an Argand diagram, as follows:
Draw a set of co-ordinate axes, and mark on the points 1 and -1 on the real axis, calling them A and B, respectively.
Mark a point z in the first quadrant somewhere near the imaginary axis, and about 2 units up (it could be anywhere, but this will be about right for the diagram to look reasonable), call this point Z.
Draw the ray from 1 through z - this is the vector z - 1, and mark it as such. Mark the angle between this vector and the positive direction of the real axis as alpha.
Draw the ray from -1 through z - this is the vector z + 1, so mark it as such. Mark the angle between this vector and the positive direction of the real axis as beta.

alpha and beta are, respectively, arg (z - 1) and arg(z + 1), and so the required locus is alpha - beta = pi / 4

The angle between the vectors is also alpha - beta, and so this locus is really saying that angle AZB is pi / 4. It follows from circle geometry that Z is any point above the real axis on a circle of which AB is a chord. Since the centre of any circle lies on the perpendicular bisector of any chord, the centre C must lie somewhere on the imaginary axis. Mark a point C on this axis, such that angle ACB is a right angle (remember that the angle at the centre is twice the angle at the circumference, and so angle ACB = 2 * angle AZB = 2 * pi / 4 = pi / 2).

Simple trigonometry should show that C lies one unit up from AB, and hence C is at i.

Thus z lies on the circle |z - i| = sqrt(2), subject to Im(z) > 0, or in cartesian form, the locus of z is
x^2 + (y - 1)^2 = 2, subject to y > 0.

 

[wogboy]

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

let z = r*(cos@ + isin@)
(|z| - iz)/(|z| + iz)
= {r - i*r(cos@ + isin@)}/{r + i*r(cos@ + isin@)}
= {1 - i*(cos@ + isin(@)}/{1 - i*(cos@ + isin@)}
= {1 - [cos(@+pi/2) + isin(@+pi/2)]}/{1 + [cos(@+pi/2) + isin(@+pi/2)]}

(multiplying by i is equivalent to adding
pi/2 to the argument)

= {1 - (-sin@ + icos@)}/{1 + (-sin@ + icos@)}
= {(1 + sin@) - icos@}/{(1 - sin@) + icos@}
= {(1 + sin@) - icos@}*{(1-sin@) - icos@}/{(1-sin@)^2 + cos^2(@)}
= {(1 + sin@)*(1-sin@) - i(cos@ - sin@*cos@) - i(cos@ + sin@*cos@) - cos^2(@)}/{2 - 2*sin@}
= {1 - sin^2(@) - cos^2(@) + i*(sin@*cos@ - cos@ - sin@*cos@ - cos@)}/{2 - 2*sin@}
= {-2i * cos@}/{2 - 2*sin@}
= -i * cos@/(1-sin@)
= -i * {cos@*(1 + sin@)}/{1 - sin^2(@)}
= -i * cos@ * (1 + sin@) / cos^2(@)
= -i * sec@ * (1 + sin@)
= -i * (sec@ + tan@) (as required)

 

[CM_Tutor]

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

Hanyway, If you can do the last one, you are a genius. If you aren't a genius, don't even attempt it.

My question, apart from the solutions to the above questions, is should I bother doing this exercise? Or is it a bit too high for HSC standards?

This one is hard, because the algebra is messy, but you still should be able to make some progress with it.

We are given that z = r(cos@ + i sin@). For convenience, I am going to define a new complex number w as z / r, and so z = rw - this is because |w| = 1. Putting this into our expression, we get:

(|z| - iz) / (|z| + iz) = (|rw| - irw) / (|rw| + irw) = r(|w| - iw) / (|w| + iw) = (1 - iw) / (1 + iw), on cancelling r and using |w| = 1.

Now, it gets messy. We know w = cos@ + isin@, and so iw = -sin@ + icos@, and thus we have:
(1 - iw) / (1 + iw) = (1 + sin@ - icos@) / (1 - sin@ + icos@), and we need to realise the denominator, giving:
(1 + sin@ - icos@)(1 - sin@ - icos@) / (1 - sin@ + icos@)(1 - sin@ - icos@)
= [(1 - icos@)^2 - (sin@)^2] / [(1 - sin@)^2 - (icos@)^2]
= [1 - 2icos@ - cos^2(@) - sin^2(@)] / [1 - 2sin@ + sin^2(@) + cos^2(@)]
= {1 - [sin^2(@) + cos^2(@)] - 2icos@} / {1 + [sin^2(@) + cos^2(@)] - 2sin@}
= (1 - 1 - 2icos@) / (1 + 1 - 2sin@), noting that sin^2(@) + cos^2(@) = 1
= -2icos@ / 2(1 - sin@)
= -icos@ / (1 - sin@)
= -icos@(1 + sin@) / (1 - sin@)(1 + sin@)
= -icos@(1 + sin@) / [1 - sin^2(@)]
= -icos@(1 + sin@) / cos^2(@), noting that sin^2(@) + cos^2(@) = 1
= -i(1 + sin@) / cos@
= -i[(1 / cos@) + (sin @ / cos@)]
= -i(sec@ + tan@), as required.

 

[OLDMAN]

________________________________________________
McLake *Brain explosion*

Let Keypad, turtle, OLDMAN or spice girl answer it ...
_______________________________________________

Did someone call my name?

Question 4 could also be approached geometrically.

Treat |z| as a real complex number in the argand plane. Now, |z|-iz and |z|+iz will be two points on the plane with midpoint |z|. Distances from |z| to O, |z|-iz, and |z|+iz are all equal to |z|. Thus these three points lie on a circle radius |z|, and vectors |z|-iz and |z|+iz include a 90 degree angle. There will be a couple of isoceles angles... and you will find that tan(45+@/2) will be the ratio | (|z|-iz)/(|z|+iz) |, which is equal to sec@+tan@.

 

[CM_Tutor]

Oldman, that is a really elegant geometric approach. Have you seen this before, or did you just see that that would work, and if you did, what tipped you off?

 

[Grey Council]

Thank you, to all those who replied.

oh, don't you worry about Oldman, that guy is a berloody genius. An absolute ocean of knowledge. Go and have a look at his credentials. Better yet, try and search for the questions he put up just before the hsc extension 2 maths exam last year. :-O

And shit me dead, fella's. Are those questions hard or what?!? I mean, the exercise before was relatively simple, and I did all but the last question with ease. And then suddenly it plunges me into these types of questions.

I mean, the first question IS actually the first question in the exercise (exercise has like 30 questions :-\ ). Absolutely blew me over.

So the verdict by the jury is I should proceed with the questions. Hokay, i'll start soon. But i'll prolly end up asking most, if not all, questions here.

EDIT
WHOA! Oldman, very elegant solution. very. lol, if i sat there thinking for a thousand years, I doubt i'd have thought of it. I hope Keypad is taking notes.

PS So all those people who solved the question, ie Tutor, that smooth fella and Oldman are geniuses. lol, looks as if OLDMAN is a genius amongst geniuses. lol

 

[Grey Council]

actually, not that hard. I should have been able to do the second and third question, especially the second one. But someone asked me to post up a few questions, so I chose at random and posted up without really thinking about it (i have the questions on computer, so i copied pasted).

I won't look at the solution to all of them just yet (except the fourth question, which I really really really doubt I would have been able to do), i'll give em all a go first.
So please, don't delete any posts in this thread. ta

 

[OLDMAN]

___________________________________________________
CM_Tutor:
Oldman, that is a really elegant geometric approach. Have you seen this before, or did you just see that that would work, and if you did, what tipped you off?
__________________________________________________

I have been worrying about students, lately, in relation to whether they really grasp the concept of a "floating vector". So lets say I have been quite obsessed by the concept, and this question came along. To jump from a vector, then to a point in the argand plane could be hard to teach to a beginner, or to a student who thinks he knows but doesn't know.

 

[CM_Tutor]

Grey Council (and everyone else), here's another one for you to think about...

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

Oldman, I'd be interested in a geometric way of finding this centre if you can see one.

 

[Grey Council]

how would I sketch 1/z?
zbar / |z|²
hrm

 

[CM_Tutor]

That's a good start. You know z is somewhere in the first quadant, but outside the unit circle. Now, where would z(bar) be? Well, start by asking yourself what would be its argument? its modulus? If your not sure, take a value for z, and try it. I'll give you some more help if you need it ...

 

[Grey Council]

hokay, so 1, -1, z anywhere in first quadrant, 1/z is inside unit circle with the negative argument of z.

 

[CM_Tutor]

All true, and the bigger the value of |z|, the closer 1 / z is to the origin.

 

[Grey Council]

i admit defeat. I am bamboozled.

Way I see it is this:
Arg z = theta
arg(z-1) = the angle the line from z to negative one makes.

so how does theta minus arg (z-1) equal to CAD? :S

PS Can you confirm that you got the email?

 

[CM_Tutor]

Hint: Vectors. Remember the interpretation of question 2 earlier in this thread, and what was done there with arguments. So, start by find an expression for the vector CA in terms of z, then vector DA ...

This is a question that is screaming out for a geometric interpretation - there probably is an algebraic approach, but you so don't want to go there if you can avoid it.