Turn on thread page Beta
 You are Here: Home >< Maths

# FP3 - Elementary transformations in the complex plane watch

1. I've been at this question for half and hour and I keep on getting the same answer (u+1)2 +v2=0 Can someone please do it for me, thank you.

w = (z + i)/(z - i), z!=i

show that as z moves along the real axis, w moves along a circle centre O and radius 1.
2. I don't know the method you tried using, but here's how I tried to do it:

Thus, considering the real and imaginary part of w, and letting x denote the real axis, and y the imaginary axis, we get:

(1)
(2)

Then I simply brute-forced this, since we know we want , and (or something along those lines). Thus we get:

(3) and
(4)

Now, I'm not sure if you've come across the substitution before (it's quite useful in integrating certain integrals involving sin and cos, such as ) - there's some information about it on wikipedia if you haven't met it yet.

Basically, you can use this substitution to write sin and cos in terms of . I presume you know the formula; thus, drawing a right-angled
triangle with one of the acute angles being , and knowing that (with ) and using the fact that , we can define the opposite = , and the adjacent = , which gives the hypotenuse = .

Therefore, using the definitions of sin and cos, we get:

and

with

Notice the similarity with the equations (3) and (4) above? That's what made me think of the substitution.

Thus, returning to equations (1) and (2), and letting , we get:
and

and so w moves along a circle centre O and radius 1, as varies from - to (Note that , rather than , as we had originally supposed) .

Notice that as varies from - to , varies from - to , so z varies from - to - that is, as varies from - to , z varies along the whole of the real axis, as desired.
3. Wow thanks a ton, that's honestly the most helpful post I've ever seen on these forums. Cheers.
4. A slightly shorter method:

As fatuous_philomath said:

w = [(z^2 - 1) + 2iz]/(z^2 +1)

we want to know the behaviour of w as z moves along the real axis ie z = x (strictly real).

w = [(x^2 - 1) + 2ix]/(x^2 +1)

now consider w = u + iv

then u = (x^2 - 1)/(x^2 +1)

and v = 2x/(x^2 +1)

notice that u^2 + v^2 = [(x^2 - 1)^2 + 4x^2]/[(x^2 + 1)^2]

= [x^4 - 2x^2 +4x^2 + 1]/[(x^2 + 1)^2]

= 1

which is the equation of a circle. So as x varies (ie as z moves along the real axis) w moves along the circle u^2 + v^2 = 1 ie centre 0, radius 1.

EDIT: v shouldn't have had an i in there...whoops.
5. (Original post by Datura)
Wow thanks a ton, that's honestly the most helpful post I've ever seen on these forums. Cheers.
You're welcome.

(Original post by wacabac)
... snip ...
Thanks, I was wondering where Datura had got u and v from; I hadn't done transformations in the complex plane before. Now I see.
6. Ok, there is a different way to do this. There are three really useful things to remember when dealing with complex numbers.

(1) |z|2 = zz* (where z* is the conjugate)
(2) Re(z) = (z+z*)/2
(3) Im(z) = (z-z*)/2i

Now onto this proof.

Point 1: if z is real, then |w|=1
In other words, for real z, w is on the unit circle centred at 0. Note that if z is real then z = z* and therefore

so |w|=1 when z is real, as required.

Point 2: if |w|=1, then there is a real z which maps to w
Why is this important? We have shown that all points on the real axis end up on the unit circle, but we don't know that they "fill up" the unit circle. It could be that the real axis only gives us a semi circle of points, we have to check this isn't the case.

Now this map has an inverse, given by z = i(w+1)/(w-1) and I claim that when |w|=1 (ie. w is on the unit circle) this formula gives a real value for z. This is what we want because it shows that if |w|=1, then there is a real z which maps to w. So I want to show that when |w|=1, the imaginary value of z must be 0. Now I have missed out all the intermediate calculation, but you get (the squigily thing followed by z at the start is meant to be Im(z) but it's coming out funny)

so clearly when |w|=1, Im(z) = 0 so z is real.

Now in an exam you would probably only have to provide point 1, and then claim that was a proof, but technically you also need point 2. Let me know if you don't understand any of this.
7. wow again thanks, these are very different approches from what I've been taught so its really interesting. I assume your at uni right?
8. If you were referring to me...

Nope, I'm in Year 13. I'm hoping to do Mathematics at university next year though.
9. (Original post by Datura)
wow again thanks, these are very different approches from what I've been taught so its really interesting. I assume your at uni right?
Yeah I'm a second year mathematician, but the methods in my post are within the a-level course. You get very used to dealing with complex numbers when you do pure maths at uni, and remembering those three identities can really help , in particular |z|2 = zz*.
10. (Original post by mikesgt2)
Ok, there is a different way to do this. There are three really useful things to remember when dealing with complex numbers.

(snip)

Point 1: if z is real, then |w|=1

(snip)

Point 2: if |w|=1, then there is a real z which maps to w

(snip)

Now in an exam you would probably only have to provide point 1, and then claim that was a proof, but technically you also need point 2. Let me know if you don't understand any of this.
A similar proof in overall structure but using slightly different identities:

Point 1: |w| = 1

Noting that z can we written as u, where u is real:
Note

as required.

Point 2: arg w could be anything from to

Using some standard results:

Clearly a full range of values for u is available, so arctan of them can be anything from to [tex]\frac{\pi}{2}[tex], so the arg of w can be all values from plus to minus pi, thus giving a circle.

Ben
11. (Original post by Sculler)

I'm probably wrong but should the last line say
?

Also, I have one more question. I can do the first part fine but not the second part.

A transformation from the z-plane to the w-place is given by

Show that under this transformation the line Im(z)=0.5 is mapped to the circle with equation |w|=1. (Up to here is fine)

Hence, or otherwise find, in the form where a,b,c,d are elements of the complex numbers, the transformation that maps the line Im(z)=0.5 to the circle centre (3,-i) and radius 2.

I can't see where the 'Hence, or otherwise' bit comes in so there must be some relation I'm not aware of. Thanks for any suggestions.
12. (Original post by Datura)
I'm probably wrong but should the last line say
?
Nope, Sculler's correct.

...with the -1 occuring because .
13. (Original post by fatuous_philomath)
Nope, Sculler's correct.

...with the -1 occuring because .
I am in fact wrong, but in a different way; from the second line onward the z should be written u to be clear we are considering the real and imaginary components of z+i or z-i, and noting z has no imaginary component. But otherwise yes, it should be -1 because i^2 = -1 as you said.
14. (Original post by Datura)
A transformation from the z-plane to the w-place is given by

Show that under this transformation the line Im(z)=0.5 is mapped to the circle with equation |w|=1. (Up to here is fine)

Hence, or otherwise find, in the form where a,b,c,d are elements of the complex numbers, the transformation that maps the line Im(z)=0.5 to the circle centre (3,-i) and radius 2.

I can't see where the 'Hence, or otherwise' bit comes in so there must be some relation I'm not aware of. Thanks for any suggestions.
Okay, let's define a complex number t = 2w+3-i; when w describes a circle radius 1 and centre (0,0) then this is the circle described. Invert this to see . This maps the circle of radius 2 cetnre (3,-1) to a unit circle about the origin. But:

This maps a unit circle about the origin to the required line. Combining the two:

This might well be the required transformation. However, given I did this straight into tex code rather than working it out on paper there may well be a mistake somewhere. Ah well.

Edit: Left out some z's at the end

Ben
15. Ben, you seem to have read the question incorrectly (I think...).

The transformation maps the circle of radius 2 centre (3, -1) to the line Im=(0.5). To find the transformation mapping the line Im=(0.5) to the circle of radius 2 centre (3, -1), which is what the question asks, we want t in terms of z:

Or simply combining the transformation with the transformation (which is probably the expected way of answering the question):

as before.
16. (Original post by fatuous_philomath)
Ben, you seem to have read the question incorrectly (I think...).

The transformation maps the circle of radius 2 centre (3, -1) to the line Im=(0.5). To find the transformation mapping the line Im=(0.5) to the circle of radius 2 centre (3, -1), which is what the question asks, we want t in terms of z:

Or simply combining the transformation with the transformation (which is probably the expected way of answering the question):

as before.
Ok cool thats the answer.

(Original post by fatuous_philomath)
Nope, Sculler's correct.

...with the -1 occuring because .
Ok I understand that but I'm still a bit unsure about finding the modulus of a complex number. I've just assumed that we ignore the i, like we do in vectors (and since |i|=1), and just square the coefficient of the i value. In my text book its has examples like this |(x-3)+iy| --> (x-3)2 + y2. Wouldn't this also imply [(x-3)2 + y2]0.5 Which is the same sort of form as you've written??

Thanks for taken the time out to do these questions, its appreciated.
17. (Original post by fatuous_philomath)
Ben, you seem to have read the question incorrectly (I think...).
Yep, I read it the other way round, so tried to get from the circle to the line. Sorry, doing that made it more complicated than it needed to be.
18. (Original post by Datura)
Ok I understand that but I'm still a bit unsure about finding the modulus of a complex number. I've just assumed that we ignore the i, like we do in vectors (and since |i|=1), and just square the coefficient of the i value. In my text book its has examples like this |(x-3)+iy| --> (x-3)2 + y2. Wouldn't this also imply [(x-3)2 + y2]0.5 Which is the same sort of form as you've written??

Thanks for taken the time out to do these questions, its appreciated.
I seem to be doing a lot of backpedalling atm (probably a result of doing these questions too late at night) but I retract what I claimed: you're right. You do only use the coefficient of the imaginary part, which in this case is + 1, and squared gives 1 as you say.
I think the confusion was caused by me missing a step, if I were to have written
Then sepreated z = u + iv, where u,v are real and noting v=0:

from which we clearly get as you claim.
Sorry for causing the confusion.

Ben
19. (Original post by Datura)
Ok I understand that but I'm still a bit unsure about finding the modulus of a complex number. I've just assumed that we ignore the i, like we do in vectors (and since |i|=1), and just square the coefficient of the i value. In my text book its has examples like this |(x-3)+iy| --> (x-3)2 + y2. Wouldn't this also imply [(x-3)2 + y2]0.5 Which is the same sort of form as you've written??

Thanks for taken the time out to do these questions, its appreciated.
Woops.

Thanks for correcting us; it's better making stupid mistakes on a student forum rather than in proper exams, and because of this thread I'm (slowly) remembering the complex number stuff I've forgotten.
20. ah good, I'm glad I can help out for once!

Turn on thread page Beta
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: June 14, 2006
Today on TSR

### Which business legend are you?

We have the answer...

### University open days

• University of Bradford
Wed, 21 Nov '18
• Buckinghamshire New University
Wed, 21 Nov '18
• Heriot-Watt University
Wed, 21 Nov '18
Poll
Useful resources

## Make your revision easier

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

Can you help? Study help unanswered threads

## Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE