I've been at this question for half and hour and I keep on getting the same answer (u+1)^{2} +v^{2}=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.

Datura
 Follow
 0 followers
 1 badge
 Send a private message to Datura
 Thread Starter
Offline1ReputationRep: Follow
 1
 11062006 17:21

 Follow
 2
 11062006 19:33
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 bruteforced 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 rightangled
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. 
Datura
 Follow
 0 followers
 1 badge
 Send a private message to Datura
 Thread Starter
Offline1ReputationRep: Follow
 3
 11062006 21:08
Wow thanks a ton, that's honestly the most helpful post I've ever seen on these forums. Cheers.

 Follow
 4
 11062006 21:29
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. 
 Follow
 5
 11062006 21:40
(Original post by Datura)
Wow thanks a ton, that's honestly the most helpful post I've ever seen on these forums. Cheers.
(Original post by wacabac)
... snip ... 
mikesgt2
 Follow
 0 followers
 0 badges
 Send a private message to mikesgt2
 Visit mikesgt2's homepage!
Offline0ReputationRep: Follow
 6
 11062006 21:42
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) = (zz*)/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)/(w1) 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. 
Datura
 Follow
 0 followers
 1 badge
 Send a private message to Datura
 Thread Starter
Offline1ReputationRep: Follow
 7
 11062006 23:26
wow again thanks, these are very different approches from what I've been taught so its really interesting. I assume your at uni right?

 Follow
 8
 11062006 23:39
If you were referring to me...
Nope, I'm in Year 13. I'm hoping to do Mathematics at university next year though. 
mikesgt2
 Follow
 0 followers
 0 badges
 Send a private message to mikesgt2
 Visit mikesgt2's homepage!
Offline0ReputationRep: Follow
 9
 13062006 00:13
(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? 
 Follow
 10
 13062006 10:31
(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.
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 
Datura
 Follow
 0 followers
 1 badge
 Send a private message to Datura
 Thread Starter
Offline1ReputationRep: Follow
 11
 13062006 19:31
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 zplane to the wplace 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. 
 Follow
 12
 13062006 21:35

 Follow
 13
 13062006 23:43
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 zi, and noting z has no imaginary component. But otherwise yes, it should be 1 because i^2 = 1 as you said.

 Follow
 14
 13062006 23:57
(Original post by Datura)
A transformation from the zplane to the wplace 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.
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 
 Follow
 15
 14062006 02:24
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. 
Datura
 Follow
 0 followers
 1 badge
 Send a private message to Datura
 Thread Starter
Offline1ReputationRep: Follow
 16
 14062006 11:43
(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 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 (x3)+iy > (x3)^{2} + y^{2}. Wouldn't this also imply [(x3)^{2} + y^{2}]^{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. 
 Follow
 17
 14062006 13:48
(Original post by fatuous_philomath)
Ben, you seem to have read the question incorrectly (I think...). 
 Follow
 18
 14062006 13:56
(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 (x3)+iy > (x3)^{2} + y^{2}. Wouldn't this also imply [(x3)^{2} + y^{2}]^{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 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 
 Follow
 19
 14062006 14:21
(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 (x3)+iy > (x3)^{2} + y^{2}. Wouldn't this also imply [(x3)^{2} + y^{2}]^{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.
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. 
Datura
 Follow
 0 followers
 1 badge
 Send a private message to Datura
 Thread Starter
Offline1ReputationRep: Follow
 20
 14062006 17:54
ah good, I'm glad I can help out for once!
Related discussions
 The hard integral thread.
 A Summer of Maths 2013
 Mega A Level Maths Thread MKIII.
 Mega A Level Maths Thread  Mark IV
 The Proof is Trivial!
 MAT Prep Thread 2017 [2nd November 2017]
 Official TSR Mathematical Society
 Year 13 Maths Help Thread
 Year 12 Maths Help Thread
 Cambridge Natural Sciences (NatSci) Students and ...
Related university courses

Cymraeg/Mathematics
Aberystwyth University

Economics and Mathematics
University of Derby

Mathematics
Queen Mary University of London

Mathematics (3 years)
Durham University

Mathematics (3 years)
Durham University

Mathematics (major with a minor)
University of Leicester

Mathematics with Business and Management
Aberystwyth University

Mathematics with Ocean and Climate Sciences
University of Liverpool

Mathematics with Physics (Including Placement Year)
University of Essex

Modern Languages with Mathematics
Royal Holloway, University of London
see more
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.
 Notnek
 charco
 Mr M
 Changing Skies
 F1's Finest
 RDKGames
 davros
 Gingerbread101
 Kvothe the Arcane
 TeeEff
 Protostar
 TheConfusedMedic
 nisha.sri
 claireestelle
 Doonesbury
 furryface12
 Amefish
 Lemur14
 brainzistheword
 Quirky Object
 TheAnxiousSloth
 EstelOfTheEyrie
 CoffeeAndPolitics
 Labrador99
 EmilySarah00
 thekidwhogames
 entertainmyfaith
 Eimmanuel
 Toastiekid
 CinnamonSmol
 Qer
 The Empire Odyssey
 RedGiant
 Sinnoh