Hi Ruud I have only just seen your post today because I always look at the bottom of the blog for new posts. Where is your solution so that I can have a look at it? Vasco
Thank you for posting these up. I am taking a Intro to Complex Analysis class using this book and knowing the answer and how to get to them has really helped to do my homework and to understand the information. Keep up the work! Thanks again
Thanks for the encouragement Amber. If you find any mistakes or need any clarifications, please let me know and I'll make corrections. Remember, the answers I have posted are my own and have not been checked, so a critical approach to them is good.
Hello again. I have been working on an assignment for my class and I was wondering if you could do number 12 and 15 from chapter 2.I would like to have something to compare to and see if I am on the right track. Also, I would be very grateful if you could post it before Thursday but if not it is fine. Thank you.
While doing exercise 38 chapter 2 I noticed a couple of typing errors in the solution to exercise 37 chapter 2 in part(iii), which I have now corrected. These errors did not affect the result. I have re-posted the corrected version today.
I am typing up exercise 13 chapter 8 today. It is one of those exercises which requires a lot of typing and diagrams. I hope to finish it today, but I can't guarantee it. As you may have realised I am located in the UK, so I am 7 hours ahead of anyone in Wyoming USA. I am typing this at 7:10 am.
Just found a small mistake in my solution to Exercise 13 chapter 8. It is on the final line of part(ii) at the bottom of page 4 in my document. This has now been corrected on the website and so if you download again you will get the corrected version.
I suddenly realised that in exercise 6 chapter 8, m does not have to be an integer. I have changed the proof slightly so that m can now be any real number and I have also replaced the original version on the web site.
Hi I am glad you find my solutions useful. My aim is to eventually publish solutions to all the exercises. If you want to publish your own solutions, I would prefer that you did this on your own site. What I do is upload my solutions to Google Drive. I created my web site using Google and then provide links from there to the solutions on Google drive. All of these facilities are free. If you decide to do this, then let me have a link to your site and I will put a reference to it on my site. I haven't really skipped chapters 6 and 7, it's just that I had a request on this blog to do some of the exercises from chapter 8 and so I did those exercises first. I wish you all the best as you work through the book. If you would like me to post my solution to a particular exercise, let me know. Vasco
Sorry I didn't manage to post exercise 4 chapter 9 yesterday. I came across a couple of subtleties that took me a while to sort out. I am now editing the file so that it is in a fit state to post. Should be later today. I've heard that somewhere before. Fingers crossed.
Uploaded a corrected version of exercise 1 chapter 9. Nothing serious just a typing error. I used theta when it should have been t. Proof is unaltered.
If we write \[CurlyPhi] = b(t + \[Tau] ) then Z(t + \[Tau] ) = e^((a/b)\[CurlyPhi]) e^i\[CurlyPhi] and so the polar form of Z(t + \[Tau] ) = e^((a/b)\[CurlyPhi]) which is the same as the polar form of Z(t) obtained in exercise 27, ..."
Should this be "and so the polar form is |Z(t + \[Tau] )| = r = e^((a/b)\[CurlyPhi]) " ?
I have revised exercise 4 chapter 4. Not because I think it is wrong, but because I wanted to include a higher quality diagram and make the explanation clearer (I hope). Also I am about to publish a solution to exercise 16 chapter 5, and the hint given for solving part(ii) refers to exercise 4 chapter 4, and so I thought it would be a good idea to make sure it was easy to understand. I re-read my solution and decided it would be a good idea to revise it.
I found a typo in exercise 40 chapter 1 part (ii) on the second line from the bottom of page 2: "an anticlockwise" should be "a clockwise". I have uploaded a corrected version.
Hi Gary Ch 1, Ex. 40, Fig 1: +/- i|p|e^{phi} --> +/- i|pi|e^{i phi} I agree that e^{phi} should be e^{i phi}, but I don't agree that |p| should be |pi|. Let me know what you think.
Ch 1, Ex. 40, (v), line 8: +e^{Pi/4} --> + e^{i Pi/4} I agree with this one.
I will make the corrections and upload a corrected version of exercise 40 chapter 1. Thanks again. Vasco
>Ch 1, Ex. 40, Fig 1: +/- i|p|e^{phi} --> +/- i|pi|e^{i phi} I agree that e^{phi} should be e^{i phi}, but I don't agree that |p| should be |pi|. Let me know what you think.
Hi The RTR site is not mine. I just link to it from this site. However I can communicate with the owner, which I did in this case. If anyone notices a problem with the RTR site please let me know and I'll get in touch with the owner on your behalf. Vasco
Unfortunately the owner of the site has not replied yet, so I do not know what is happening. I have contacted him by email and also via Twitter. If it becomes clear that this situation is permanent then we could consider starting all over again. It would be a pity to lose all those previous solutions though. Hopefully some previous contributors would resubmit their solutions.
Mathematica 8 doesn't render the feet very well, but it gets the idea across. Plot available if desired.
Needham, Chapter 2, Exercises 7-
7. Sketch the modular surface of C(z) = (z+1)(z-1)(z+1+i). Hence, sketch the Cassinian curves |C[z]| = const., then check your answer using a computer.
This is a followup on my comment of March 25. I posed the plotting problem to Stack Exchange Mathematica and got a nice answer from Rajul Narain, who modified the second line, so the useful script is now:
Thanks Gary, that's really great! I used your link and found the really wonderful graphic. I notice that there are 3 sets of curves depicted in the graphic: the Cassinian curves, the curves at right angles to the Cassinian curves and another set. I'm not sure what the third set represents. Would it be possible to produce this picture without this third set of curves? With your permission I would like to incorporate this into my solution on my website. I would of course acknowledge you and Rajul in the text. The exercise also asks for a drawing of the Cassinian curves from above as in figures [8] anf [9] on pages 61-62 of the book. Would it be possible to produce a graphic of these as well? I do not have access to Mathematica as you know. Did you buy it or do you access it some other way? Regards Vasco
Vasco, I just saw this reply today. I'll take a look and see what I can do. I would be happy to contribute, as I am benefiting from your web site. It may be next week before I can get to it.
I purchased a home copy of Mathematica 8 for about $200 a couple of years ago.
Vasco, All the curves shown on the surface are modeling curves drawn automatically by Mathematica. I can easily redraw it as a surface without any superimposed curves.
It would also be easy to include any desired figure captions or other text in the image.
It may be possible to draw just the Cassinian curves and the orthogonals and add them to the graphic. I will have to experiment with that. I'm apprehensive that it will be tricky. I did that with the two-dimensional image and found that to draw the orthogonals I had to seek other equations than those found in the Needham text.
Vasco, All three sets of curves drawn automatically on the modular surface by Mathematica can be shown independently. There is a cassinian set and two other sets. These sets appear not to be orthogonal to the cassinian set, but they are orthogonal to each other. See for example:
Thanks Gary. Can you do one where you are looking down on the "cube" from directly above so that the picture looks like figure 8b in the book on page 61?
Vasco, I got distracted and haven't been back to this page, so I just found your question. In fact, I had already done that in the 2D figure, but I also got something like it by rotating the 3D plot. Fine tuning is possible. You can always contact me directly at gxp!alx mer!lv (remove the x's and !s and space). If I get time, I will post these on the web. I may remove them from the Dropbox to save hard drive space.
You are right that there is a typo in equation (3), the denominator should be squared. However the rest of my solution is correct. I think you must have made some algebraic errors when solving for A & B. Don't forget that (1+k^2)=(-k-i)(-k+i). Multiplying equation (3) by(1+k^2)=(-k+i)(-k-i), gives: A(-k+i)+B(-k-i)=-2k/(1+k^2) (4) Multiplying equation (2) by (-k+i) gives: A(-k+i)+B(-k+i)=(-k+i)/(1+k^2)=1/(-k-i) (5) Subtracting (5) from (4) gives -2iB=(-k-i)/(1+k^2)=1/(-k+i) and so B=-1/(2i) x 1/(-k+i) {where x means multiply} Substituting this value for B into equation 2 gives A=1/(2i) x 1/(-k-i) which are the values in my published solution. My solution is correct apart from the typo that you spotted. Thanks. I will correct it and upload the corrected version. Vasco
Hi, I haven't looked at these exercises yet. I will take a look at them and try and do as many of them as possible during the next week. I may not be able to complete them all in a week.
A question regarding Chapter 2, Section 2, Single-valued branches of a multifunction.
On pages 93-94 we are shown how to create a branch cut with a curve C, leaving a plane with the points of C removed. Then it reads “We thereby obtain on S the three branches f_1, f_2, and f_3.” How is it that one cut can create three branches? Figure [32](b) appears to show three cuts. Is it that “thereby” implies making two additional cuts?
Hi Gary Thanks for the new graphic - it looks great! The only improvement would be if you could rotate it so that the points z=-1 and z=1 are horizontal in your picture, as they are when normally drawn with real and imaginary axes. Don't draw the axes, just rotate the picture. Thanks . Please find attached here: https://drive.google.com/file/d/0B4HzA5QgnnvZUm9PUDN0NFZ2aTg/edit?usp=sharing a document which I hope will help you to understand the multifunction section. If after reading it you have any further questions, please ask. Vasco
Here is another version of the trifocal Cassinian plot with f1, f2 aligned horizontally. See if it is what you had in mind. I removed the box lines, but you might prefer them for perspective.
Thanks Gary. I have downloaded all your graphs and I will use two of them in my solution to Exercise 7 chapter 2. I like the plain Cassini curves graph with no colouring and the 3D one from the side with only 'horizontal' curves. The last one you sent is fine but I find the colouring too distracting. A simpler colouring scheme which gives the impression of going down inside the legs would be better, but the plain one I mentioned above with no colouring is fine for my purposes. Thanks again. Vasco
Glad to help. I think I will leave it at that for the time being. I'm busy trying to figure out the criteria for ascertaining whether a branch is a principle branch (Ch 2, Ex. 18 (vi).)
For your interest, here is an item based loosely on pp. 126-128. It requires the Wolfram CDF Player, which is a free download.
I failed A level maths 60 years ago - my excuse is that I tried to do it in one year at the age of 15 but have regretted ever since not having the time to pursue maths until retirement. However I bought RTR a couple of years back and have worked through that up to Chapter 7. I then bought Needham in the hope of understanding Complex Analysis but I find he says that he does not cover Riemann surfaces which have baffled me in RTR at Chapter 8. Is there some other book on Riemann surfaces I should read?
I find Needham very tough and was tempted to give up until I found your site which I have found a great help. It is difficult to relate the exercises to the text where I have got to page 16 and exercise 6. I find Needham takes a lot for granted. For instance it never occurred to me that c in exercise 6 was necessarily the centre of the circle until I read your solution. I can see how you arrive at the absolute value of c being 5 but I am not clear what that means! I thought we were dealing with a circle in the composite plane with centre at (-3,4). It then seemed to me that the maximum and minimum values of z would be measured against the imaginary axis i.e. 4+2 and 4-2. Yet you say the answer is 5+2 and 5-2 as if c is in some other plane and your diagram shows max and min values of z as not being on a vertical line through the centre. Perhaps I am just fending off alzheimers but any help in understanding this would be gratefully received!
Hi Nicolas and Gary I am on the same journey as you. I started with RTR and got side-tracked into Needham. If you look in the index to Needham and follow up the Riemann surface bits, you will find some references to books you could read on this topic. I find that because of Needham's mainly visual approach it is a good idea always to look for simple ways to solve the exercises. That's what he is looking for in my opinion. Looking at |z-c|=R made me think of circles because I know that c and R are often used as centre and radius of circles, and also this equation reads to me as: "the distance between any point z and the fixed point c is constant" so z must lie on a circle. The absolute value of c, which is 5, means that the distance of the centre of the circle from the origin is 5. We are talking of the max and min values of the absolute values of z not z itself, just to be clear. We cannot talk about maximum or minimum values for complex numbers themselves - it doesn't make sense. So it's the max and min of |z| we are looking for: Get some graph paper or just make a sketch and draw the circle centred at (-3,4) with radius 2 and you will see that the circle does not cut through the x and y axes and does not enclose the origin. Draw a line from the origin through the centre of the circle and extend it until it cuts the circle again. Choose a few points on the circle and draw a line from these various points on the circle to the origin. The lengths of these lines are the values of |z| at these points. You will soon see that if you start from the point on the circle where the line through the centre intersects it (the one nearest the origin) then as you move round the circle from this point in a clockwise or anticlockwise direction, the distance from the point where you are to the origin always increases until you reach the point on the other side of the circle where the line through the centre intersects it. At this point if you continue to move round the circle then the distance to the origin starts to decrease again until you reach the point where you started. It is clear that the 2 points of intersection with the circle, of the line through the origin and the centre of the circle,are the maximum and minimum distances of points on the circle from the origin. Since we know that the circle has a radius of 2 then then the minimum distance from the origin of a point on the circle must be the difference between the distance of the centre from the origin and the radius, and the maximum distance from the origin of a point on the circle must be the sum of the distance of the centre from the origin and the radius. This is very difficult to explain in words - it would be much easier if I were talking to you and had a diagram in my hand. Thanks for pointing out that the maximum and minimum positions for z are missing from my published solution. I will amend my solution and re-publish it. Here is how I would calculate them: Since we know that cos[arg(c)]=-3/5 and sin[arg(c)] is 4/5, and since arg(z_min)=arg(z_max)=arg(c), then z_min=|z_min| {cos[arg(c)] + i sin[arg(c)]} and z_max=|z_max| {cos[arg(c)] + i sin[arg(c)]} and so z_min=3 x -3/5 + i 3 x 4/5=-9/5 + i 12/5 and z_max=7 x -3/5 + i 7 x 4/5=-21/5 + i 28/5
I hope this helps you. If you have any more questions please ask. I would ideally like this forum to be used more than it is. I publish all my solutions but get very little feedback, except from you and Gary! Thanks Vasco
If you don’t mind someone else chiming in, would it help to know that the absolute value of c is the length of the segment from the origin of the coordinates to the centre c of the circle? The absolute value (length) |z| is measured on a line segment from the origin of the coordinates to the point z, which in this case is on the circle. It is not measured vertically unless for z = x + iy, x happens to be zero.
Gary: After writing my question I had noticed that Vasco had drawn a line from the origin to the centre and where it cuts the circle I got the values 3 and 7 by some rather laborious and roundabout trigonometry. But you have clarified for me what should have been blindingly obvious from the start and showed my calculations to be unnecessary!
However I do notice that Needham asks not only for the max and min values but also the position of z but which I presume he means the co-ordinates and therefore one does need to do the trigonometry?
In real number calculus one would get at min and max values by differentiating. Can one do the same in complex analysis?
I think my problem with Needham is that he goes too fast. I am used to more traditional text books where there are two or three examples of how to solve a particular problem followed by perhaps twenty or so exercise questions which serve to really drum the ideas into one.
Although the positions can be calculated as you suggest, I don't think this is in the spirit of what Needham is trying to get across at this point. I think he wants the reader to realise that the argument of the complex numbers c, z_min and z_max is the same and therefore since we know the sin and cosine of this argument and have previously calculated |z_min| and |z_max|, we can easily calculate z_min and z_max using z=r(cos(\theta)+isin(\theta) where theta is the argument referred to above.
It’s easy to miss the forest for the trees. Some easy geometry with similar triangles will get those coordinates. Yes, the position would refer to the coordinates as Vasco writes them, e.g. (-3, 4). I get (-9/5, 12/5) for the position of z with min |z|. Of course you could also write z = -9/5 + 12i/5. Note that the diagram for part (ii) is a bit different from the general one that Vasco shows because the centre of the circle is left of the y coordinate.
As for complex derivatives, it looks like chapters 4 and 5 will get into that.
Needham assumes a lot of background and moves fast. I have gotten stuck at several points. On the other hand, the book forces review and integration of many topics and the polar notation z = re^(i theta) provides a way of calculating and thinking about positions in two dimensions that is often more efficient and intuitive than vector notation. I even finding myself applying it to ideas that I think of as coming from linear algebra, as when I calculated the minimum position above as (|c|-r) (c/|c|), where (c/|c|) is analogous to a unit vector. It’s a number of length 1 with the same angle as c. But Vasco provides a more instructive equation.
Nicolas, I tried to improve the reasoning and correct a couple of errors in the description for problem 9 in Chapter 1.
9. Let A, B, C, D be four points on the unit circle. If A + B + C + D = 0, show that the points must form a rectangle.
We assume that the points are described with complex numbers on the Argand plane and that the circle is drawn with center at the origin. Since A, B, C, and D could be any four points, we rotate the points equally (or we rotate the axes) until the x axis is parallel to side AB. We use A = a1 + a2 i, where a1 = Re[A] = cos Arg(A) and a2 = Im[A] = sin Arg(A), etc. to represent the points.
Diagram at https://www.dropbox.com/s/fd9e1c7ia0a2chp/ch1.ex9.4pts.on.circle.jpg
If the points sum to zero, they must still sum to zero after rotation, because rotation on the circle preserves the shape. Both the real and imaginary points must sum to zero. Then we can write:
We write:
A = a1 + a2 i B = b1 + b2 i = -a1 + a2 i (because the real points of A and B—being on the same vertical line—must cancel and the imaginary points must be equal)
C = c1 + c2 i = a1 - a2 i (because, by the diagram, the imaginary points of A and C must cancel and the real points must be equal)
D = d1 + d2 i = -a1 - a2 i (because, by the diagram, both the real points and the imaginary points of A and D must cancel).
Summing these up, they add up to zero. If they do not sum to zero, the four points can not describe a rectangle.
***************
It seems that every time I try to delete and add a comment, the original comment gets restored. This is my last attempt for this comment.
Gary 1. We don't have to assume that the circle is centred at the origin. When we say in maths THE unit circle, we mean a circle of unit radius centred at the origin. When we say A unit circle we mean a circle of unit radius without specifying the centre. 3. You are assuming that the points ABCD are the vertices of a rectangle - you have to prove this.
I think I was not seeing the continuation link at the bottom of the screen. I will try to clean up the redundant comments at the first opportunity. The PC and Google browser that I am using this week does not always display the delete link. In the meantime, please feel free to delete any redundant comments.
Hi Gary I have replied to one of your posts above that I think you wanted to delete. I can delete it for you, but I wanted to check with you first. Is it the one on 14th July at 4:48pm?
Just uploaded some more solutions to exercises at end of chapter 1. More to come...
ReplyDeletepodrias subir algunas del capitulo 3
DeleteRecently added solutions to exercises 5, 6, 7, 8, 9, 14, and 30.
ReplyDeleteAdded exercise 31
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ReplyDeleteI have now completed chapter 1 exercises. Please let me know if you have any questions, corrections or constructive criticisms.
ReplyDeleteI have solved exercise 43 chapter 1 without an use of reflections. I think it is more direct and more clear. What do you think?
DeleteHi Ruud
DeleteI have only just seen your post today because I always look at the bottom of the blog for new posts. Where is your solution so that I can have a look at it?
Vasco
Added exercise 1 chapter 2
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ReplyDeleteAdded exercise 13 chapter 4. Exercise 13 for the beginning of 2013!
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ReplyDeleteAdded corrected version of exercise 14 chapter 4. Equation (1) had an 'i' in it which shouldn't have been there.
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ReplyDeletepodrias subir algunos del capitulo 3
ReplyDeletete lo agradeceria mucho
Cierto, pero se tardarà algunos dias.
Deleteno ay problema, bueno si no pasan de unos 5 dias entonces si!
ReplyDeletepor favor podrias subir los que puedas ya saves del capitulo 3
ReplyDeleteEjercicio 1 casi completado. Lo subiré hoy más tarde.
DeleteHola Marcos! debes comprender que no habÃa ya hecho los ejercicios del capitulo 3. Los estoy haciendo ahora.
DeleteVasco
Added exercise 1 chapter 3
ReplyDeleteAdded exercise 2 chapter 3
ReplyDeleteThank you for posting these up. I am taking a Intro to Complex Analysis class using this book and knowing the answer and how to get to them has really helped to do my homework and to understand the information. Keep up the work! Thanks again
ReplyDeleteThanks for the encouragement Amber. If you find any mistakes or need any clarifications, please let me know and I'll make corrections.
DeleteRemember, the answers I have posted are my own and have not been checked, so a critical approach to them is good.
Hello again. I have been working on an assignment for my class and I was wondering if you could do number 12 and 15 from chapter 2.I would like to have something to compare to and see if I am on the right track. Also, I would be very grateful if you could post it before Thursday but if not it is fine. Thank you.
ReplyDeleteHi Amber
DeleteI have posted exercise 12. I will try and post 15 later today.
Hi Amber
DeleteHave only just finished doing exercise 15 on paper. I will post tomorrow. Sorry it's late for you. I hope it's of some use to you anyway.
Added exercise 12 chapter 2
ReplyDeleteAdded exercise 15 chapter 2
ReplyDeleteThanks for posting the previous problems. They really helped. Can you also post problems 25 and 37 from chapter 2 as well. Thank you.
ReplyDeleteYes, thanks for the previous problems. They did help. I would also like problems 25 and 37 from chapter 2. Thank you.
DeleteI would greatly appreciate it if you got those two problems up by Sunday. Thank you.
DeleteAdded exercise 37 chapter 2
ReplyDeleteThank You!
DeleteI have uploaded a new version which just removes a redundant sentence from paragraph 1. The solution details are not affected
DeleteAdded exercise 25 chapter 2
ReplyDeleteLet me know if you want me to post anymore particular exercises.
ReplyDeleteThank you for posting the problems. I will let you know if there are any more I need to see.
DeleteWhile doing exercise 38 chapter 2 I noticed a couple of typing errors in the solution to exercise 37 chapter 2 in part(iii), which I have now corrected. These errors did not affect the result. I have re-posted the corrected version today.
ReplyDeleteAdded exercise 38 chapter 2
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ReplyDeleteHello again. I would appreciate it if you could post the answers to problems 1,2,6,7,and 13 from chapter 5. Thank you for you help.
ReplyDeleteWill do.
DeleteAdded exercise 1 chapter 5
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ReplyDeleteI will post exercise 13 chapter 5 tomorrow.
ReplyDeleteAdded exercise 13 chapter 5
ReplyDeleteI hope to post exercise 7 tomorrow 25th March 2013.
ReplyDeleteThank you very much!
DeleteAdded exercise 7 chapter 5.
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ReplyDeleteHello again. Can you post problems 5, 13, 18, and 24 from chapter 8? Thank you.
ReplyDeleteHi Amber
ReplyDeleteI'll have a look at them starting tomorrow.
Vasco
Added exercise 18 chapter 8
ReplyDeleteAdded exercise 5 chapter 8
ReplyDeleteAdded exercise 24 chapter 8
ReplyDeleteI am typing up exercise 13 chapter 8 today. It is one of those exercises which requires a lot of typing and diagrams. I hope to finish it today, but I can't guarantee it.
ReplyDeleteAs you may have realised I am located in the UK, so I am 7 hours ahead of anyone in Wyoming USA. I am typing this at 7:10 am.
Added exercise 13 chapter 8
ReplyDeleteJust found a small mistake in my solution to Exercise 13 chapter 8. It is on the final line of part(ii) at the bottom of page 4 in my document. This has now been corrected on the website and so if you download again you will get the corrected version.
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ReplyDeleteI suddenly realised that in exercise 6 chapter 8, m does not have to be an integer. I have changed the proof slightly so that m can now be any real number and I have also replaced the original version on the web site.
ReplyDeleteI intend to to work through Visual complex analysis this summer, and your solutions are a great resource. Keep up the good work!
ReplyDeleteIf you don't mind I'd be able to contribute some solutions as well. Let me know if you are interested.
P.S. Any reason that you skipped chapter 6-7?
Hi
DeleteI am glad you find my solutions useful. My aim is to eventually publish solutions to all the exercises.
If you want to publish your own solutions, I would prefer that you did this on your own site. What I do is upload my solutions to Google Drive. I created my web site using Google and then provide links from there to the solutions on Google drive. All of these facilities are free.
If you decide to do this, then let me have a link to your site and I will put a reference to it on my site.
I haven't really skipped chapters 6 and 7, it's just that I had a request on this blog to do some of the exercises from chapter 8 and so I did those exercises first.
I wish you all the best as you work through the book. If you would like me to post my solution to a particular exercise, let me know.
Vasco
Hi. Can you do problems 1, 3, and 4 from chapter 9 as soon as possible. I am reviewing for my final in my class and I need it soon. Thank you so much.
ReplyDeleteAdded exercise 1 chapter 9
ReplyDeleteAdded exercise 3 chapter 9. I hope to post exercise 4 chapter 9 later today.
ReplyDeleteSorry I didn't manage to post exercise 4 chapter 9 yesterday. I came across a couple of subtleties that took me a while to sort out. I am now editing the file so that it is in a fit state to post. Should be later today. I've heard that somewhere before. Fingers crossed.
ReplyDeleteAdded exercise 4 chapter 9.
ReplyDeleteUploaded a corrected version of exercise 1 chapter 9. Nothing serious just a typing error. I used theta when it should have been t. Proof is unaltered.
ReplyDeleteAdded exercise 19 chapter 4.
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ReplyDeleteAdded exercise 20 chapter 5
ReplyDeleteIt is very nice to have these for self-study in an area where few people who share these interests. Thank you for posting them.
ReplyDeleteGary in Northern California, USA
Thanks Gary - glad you find them useful. If you would like me to try and post a particular exercise, let me know.
DeleteAdded exercise 22 chapter 5
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ReplyDeleteproblem 28 (i):
ReplyDeleteIf we write \[CurlyPhi] = b(t + \[Tau] ) then Z(t + \[Tau] ) = e^((a/b)\[CurlyPhi]) e^i\[CurlyPhi] and so the polar form of Z(t + \[Tau] ) = e^((a/b)\[CurlyPhi]) which is the same as the polar form of Z(t) obtained in exercise 27, ..."
Should this be "and so the polar form is |Z(t + \[Tau] )| = r = e^((a/b)\[CurlyPhi]) " ?
Gary
Yes, you are right. The way I have written it is wrong. I will amend it and put the corrected version on the website. I will amend it as follows:
Delete"...and so the polar form of Z(t + \[Tau] ) is r=e^((a/b)\[CurlyPhi]) " , which is..."
Thanks for pointing this out.
Vasco
Added exercise 21 chapter 5
ReplyDeleteMy pleasure.
ReplyDeleteGary
Added exercise 8 chapter 5
ReplyDeleteAdded exercise 15 chapter 5
ReplyDeleteI have revised exercise 4 chapter 4. Not because I think it is wrong, but because I wanted to include a higher quality diagram and make the explanation clearer (I hope).
ReplyDeleteAlso I am about to publish a solution to exercise 16 chapter 5, and the hint given for solving part(ii) refers to exercise 4 chapter 4, and so I thought it would be a good idea to make sure it was easy to understand. I re-read my solution and decided it would be a good idea to revise it.
Typo in 34 (v) line -2: "Since ζ + ζ = Re(ζ)" should be "Since ζ + ζ = 2 Re(ζ)".
ReplyDeleteThanks. I have uploaded a corrected version of exercise 34 chapter 1.
DeleteAdded exercise 27 chapter 2
ReplyDeleteTypo, Ch 1, Ex. 36, (i) bottom line: (c-b) --> (c-a)
ReplyDeleteGary
Thanks Gary. I have uploaded a corrected version of exercise 36 chapter 1.
DeleteVasco
I found a typo in exercise 40 chapter 1 part (ii) on the second line from the bottom of page 2: "an anticlockwise" should be "a clockwise".
ReplyDeleteI have uploaded a corrected version.
Added exercise 16 chapter 5
ReplyDeleteAdded exercise 28 chapter 2
ReplyDeleteAdded exercise 29 chapter 2
ReplyDeleteAdded exercise 30 chapter 2
ReplyDeleteAdded exercise 32 chapter 2
ReplyDeleteCh 1, Ex. 40, Fig 1: +/- i|p|e^{phi} --> +/- i|pi|e^{i phi}
ReplyDeleteCh 1, Ex. 40, (v), line 8: +e^{Pi/4} --> + e^{i Pi/4}
Gary
Hi Gary
DeleteCh 1, Ex. 40, Fig 1: +/- i|p|e^{phi} --> +/- i|pi|e^{i phi}
I agree that e^{phi} should be e^{i phi}, but I don't agree that |p| should be |pi|. Let me know what you think.
Ch 1, Ex. 40, (v), line 8: +e^{Pi/4} --> + e^{i Pi/4}
I agree with this one.
I will make the corrections and upload a corrected version of exercise 40 chapter 1.
Thanks again.
Vasco
>Ch 1, Ex. 40, Fig 1: +/- i|p|e^{phi} --> +/- i|pi|e^{i phi}
DeleteI agree that e^{phi} should be e^{i phi}, but I don't agree that |p| should be |pi|. Let me know what you think.
|pi| was a typo on my part.
Gary
Ch 1, Ex. 41, (ii), e^{alpha/2} --> e^{i alpha/2} in three instances.
ReplyDeleteGary
I agree. I will make the corrections and upload a corrected version of exercise 41 chapter 1.
DeleteThanks again.
Vasco
Added exercise 40 chapter 2
ReplyDeleteAdded exercise 36 chapter 2
ReplyDeleteAdded exercise 14 chapter 2
ReplyDeleteI cannot access you site of RTR, it is apparaently closed. I left you a msg on twitter too. Plz tell me when this site be back online.
ReplyDeleteIt is back on line, Thank you :)
DeleteHi
DeleteThe RTR site is not mine. I just link to it from this site. However I can communicate with the owner, which I did in this case. If anyone notices a problem with the RTR site please let me know and I'll get in touch with the owner on your behalf.
Vasco
The RTR site seems to be down again. I have contacted the owner of the site.
ReplyDeleteUnfortunately the owner of the site has not replied yet, so I do not know what is happening. I have contacted him by email and also via Twitter. If it becomes clear that this situation is permanent then we could consider starting all over again. It would be a pity to lose all those previous solutions though. Hopefully some previous contributors would resubmit their solutions.
ReplyDeleteI have had a reply from the owner of the site and he says he will fix things this coming week.
ReplyDeleteChapter 2
ReplyDeleteMathematica 8 doesn't render the feet very well, but it gets the idea across. Plot available if desired.
Needham, Chapter 2, Exercises 7-
7. Sketch the modular surface of C(z) = (z+1)(z-1)(z+1+i). Hence, sketch the Cassinian curves |C[z]| = const., then check your answer using a computer.
c[z_]:=(z+1)(z-1)(z+1+I);
cp = ContourPlot3D[
Abs[c[x+I y]]==Abs[k^3],{x,-2.5,2},{y,-2,2},{k,0,1.75}];
Show[cp,Background->White,AxesLabel->{"x","y","k"}]
Added exercise 16 chapter 2
ReplyDeleteThis is a followup on my comment of March 25. I posed the plotting problem to Stack Exchange Mathematica and got a nice answer from Rajul Narain, who modified the second line, so the useful script is now:
ReplyDeletec[z_]:=(z+1)(z-1)(z+1+I);
Plot3D[Abs[c[x + I y]]^(1/3), {x, -2.5, 2}, {y, -2, 2}, PlotRange -> {0, 1.75},
Background -> White, AxesLabel -> {"x", "y", "k"}, ClippingStyle -> None,
BoxRatios -> 1, MeshFunctions -> {#1 &, #2 &, #3 &}, MaxRecursion -> 5]
Gary
See: http://mathematica.stackexchange.com/questions/44825/modular-surface-of-tri-focal-cassini-curve-contourplot3d-missing-feet#44825
Thanks Gary, that's really great! I used your link and found the really wonderful graphic. I notice that there are 3 sets of curves depicted in the graphic: the Cassinian curves, the curves at right angles to the Cassinian curves and another set. I'm not sure what the third set represents.
DeleteWould it be possible to produce this picture without this third set of curves? With your permission I would like to incorporate this into my solution on my website. I would of course acknowledge you and Rajul in the text. The exercise also asks for a drawing of the Cassinian curves from above as in figures [8] anf [9] on pages 61-62 of the book. Would it be possible to produce a graphic of these as well?
I do not have access to Mathematica as you know. Did you buy it or do you access it some other way?
Regards
Vasco
Vasco, I just saw this reply today. I'll take a look and see what I can do. I would be happy to contribute, as I am benefiting from your web site. It may be next week before I can get to it.
DeleteI purchased a home copy of Mathematica 8 for about $200 a couple of years ago.
Gary
Vasco, All the curves shown on the surface are modeling curves drawn automatically by Mathematica. I can easily redraw it as a surface without any superimposed curves.
Deletehttps://dl.dropboxusercontent.com/u/35393965/trifocal.cassinian.modular.surface.jpg
It would also be easy to include any desired figure captions or other text in the image.
It may be possible to draw just the Cassinian curves and the orthogonals and add them to the graphic. I will have to experiment with that. I'm apprehensive that it will be tricky. I did that with the two-dimensional image and found that to draw the orthogonals I had to seek other equations than those found in the Needham text.
Gary
Vasco, All three sets of curves drawn automatically on the modular surface by Mathematica can be shown independently. There is a cassinian set and two other sets. These sets appear not to be orthogonal to the cassinian set, but they are orthogonal to each other. See for example:
Deletehttps://dl.dropboxusercontent.com/u/35393965/trifocal.cassini.modular.surface.cas.mesh.jpg
https://dl.dropboxusercontent.com/u/35393965/trifocal.cassini.cas.mesh.ortho.mesh.jpg
https://dl.dropboxusercontent.com/u/35393965/trifocal.cassini.ortho.ortho.jpg
Thanks Gary. Can you do one where you are looking down on the "cube" from directly above so that the picture looks like figure 8b in the book on page 61?
DeleteVasco, I got distracted and haven't been back to this page, so I just found your question. In fact, I had already done that in the 2D figure, but I also got something like it by rotating the 3D plot. Fine tuning is possible. You can always contact me directly at gxp!alx mer!lv (remove the x's and !s and space). If I get time, I will post these on the web. I may remove them from the Dropbox to save hard drive space.
Deletehttps://www.dropbox.com/s/eywulhzjw15di2j/needham.ch2.exercies7.trifocal.cassinian.gif
https://www.dropbox.com/s/j8h1v6trbnacueu/needham.ch2.exercies7.trifocal.cassinian.top.view.gif
Gary
That address is at gmail.com.
DeleteOne more color plot with only Cassini lines
Deletehttps://www.dropbox.com/s/04w3cw5rznlnczv/needham.ch2.exercise7.trifocal.cassinian.top.view.2.gif
Gary
At long last the RTR web site www.roadtoreality.info is working again.
ReplyDeleteChapter 2, Exercise 9, part (iv), at (3), shouldn't c_1 have have the denominator squared? I.e. c_1 = -2k/(1+k^2)^2.
ReplyDeleteIf so, then I think
A = (i+k)/2i(1+k^2) (or -(-i-k)/2i(1+k^2)
and
B = (-k+i)/2i(1+k^2)
An extra round of partial fractions is required to get these.
Gary
You are right that there is a typo in equation (3), the denominator should be squared. However the rest of my solution is correct. I think you must have made some algebraic errors when solving for A & B. Don't forget that (1+k^2)=(-k-i)(-k+i).
DeleteMultiplying equation (3) by(1+k^2)=(-k+i)(-k-i), gives:
A(-k+i)+B(-k-i)=-2k/(1+k^2) (4)
Multiplying equation (2) by (-k+i) gives:
A(-k+i)+B(-k+i)=(-k+i)/(1+k^2)=1/(-k-i) (5)
Subtracting (5) from (4) gives
-2iB=(-k-i)/(1+k^2)=1/(-k+i)
and so B=-1/(2i) x 1/(-k+i) {where x means multiply}
Substituting this value for B into equation 2 gives
A=1/(2i) x 1/(-k-i)
which are the values in my published solution.
My solution is correct apart from the typo that you spotted.
Thanks. I will correct it and upload the corrected version.
Vasco
Yes, I made it unnecessarily complicated.
DeleteMy exam is in a week and I still haven't solved these problems from Ch. 3: Q 5, 10, 11,
ReplyDelete20, 21, 24, 25. Would greatly appreciate your input!
Hi,
DeleteI haven't looked at these exercises yet. I will take a look at them and try and do as many of them as possible during the next week. I may not be able to complete them all in a week.
Added exercise 5 chapter 3
ReplyDeleteAdded exercise 20 chapter 3, but only parts (i) and (ii). I will add part (iii) at a later date.
ReplyDeleteSorry I wasn't able to complete them all in one week.
ReplyDeleteA question regarding Chapter 2, Section 2, Single-valued branches of a multifunction.
ReplyDeleteOn pages 93-94 we are shown how to create a branch cut with a curve C, leaving a plane with the points of C removed. Then it reads “We thereby obtain on S the three branches f_1, f_2, and f_3.” How is it that one cut can create three branches? Figure [32](b) appears to show three cuts. Is it that “thereby” implies making two additional cuts?
Gary
Hi Gary
DeleteI went away on holiday on 1st June and have just returned and read your post. I will post a full answer as soon as I can.
Hi Gary
DeleteThanks for the new graphic - it looks great! The only improvement would be if you could rotate it so that the points z=-1 and z=1 are horizontal in your picture, as they are when normally drawn with real and imaginary axes. Don't draw the axes, just rotate the picture. Thanks
.
Please find attached here:
https://drive.google.com/file/d/0B4HzA5QgnnvZUm9PUDN0NFZ2aTg/edit?usp=sharing
a document which I hope will help you to understand the multifunction section. If after reading it you have any further questions, please ask.
Vasco
See also response of June 2.
ReplyDeleteGary
That helped. I think I have it now. I will reorient the Cassini plot.
ReplyDeleteGary
Here is another version of the trifocal Cassinian plot with f1, f2 aligned horizontally. See if it is what you had in mind. I removed the box lines, but you might prefer them for perspective.
ReplyDeletehttps://www.dropbox.com/s/te3psoxgi9m1syx/trifocal.cassinian.mod.surf.topview.gif
Thanks Gary. I have downloaded all your graphs and I will use two of them in my solution to Exercise 7 chapter 2. I like the plain Cassini curves graph with no colouring and the 3D one from the side with only 'horizontal' curves. The last one you sent is fine but I find the colouring too distracting. A simpler colouring scheme which gives the impression of going down inside the legs would be better, but the plain one I mentioned above with no colouring is fine for my purposes. Thanks again.
DeleteVasco
Vasco,
ReplyDeleteGlad to help. I think I will leave it at that for the time being. I'm busy trying to figure out the criteria for ascertaining whether a branch is a principle branch (Ch 2, Ex. 18 (vi).)
For your interest, here is an item based loosely on pp. 126-128. It requires the Wolfram CDF Player, which is a free download.
https://faculty.unlv.edu/gbp/mma/VisualComplexAnalysis/complexInversion.html
The comment process is acting differently. It is not putting up the captcha or posting comments unless you use the preview.
Gary
I failed A level maths 60 years ago - my excuse is that I tried to do it in one year at the age of 15 but have regretted ever since not having the time to pursue maths until retirement. However I bought RTR a couple of years back and have worked through that up to Chapter 7. I then bought Needham in the hope of understanding Complex Analysis but I find he says that he does not cover Riemann surfaces which have baffled me in RTR at Chapter 8. Is there some other book on Riemann surfaces I should read?
ReplyDeleteI find Needham very tough and was tempted to give up until I found your site which I have found a great help. It is difficult to relate the exercises to the text where I have got to page 16 and exercise 6. I find Needham takes a lot for granted. For instance it never occurred to me that c in exercise 6 was necessarily the centre of the circle until I read your solution. I can see how you arrive at the absolute value of c being 5 but I am not clear what that means! I thought we were dealing with a circle in the composite plane with centre at (-3,4). It then seemed to me that the maximum and minimum values of z would be measured against the imaginary axis i.e. 4+2 and 4-2. Yet you say the answer is 5+2 and 5-2 as if c is in some other plane and your diagram shows max and min values of z as not being on a vertical line through the centre. Perhaps I am just fending off alzheimers but any help in understanding this would be gratefully received!
Hi Nicolas and Gary
DeleteI am on the same journey as you. I started with RTR and got side-tracked into Needham. If you look in the index to Needham and follow up the Riemann surface bits, you will find some references to books you could read on this topic.
I find that because of Needham's mainly visual approach it is a good idea always to look for simple ways to solve the exercises. That's what he is looking for in my opinion. Looking at |z-c|=R made me think of circles because I know that c and R are often used as centre and radius of circles, and also this equation reads to me as:
"the distance between any point z and the fixed point c is constant" so z must lie on a circle.
The absolute value of c, which is 5, means that the distance of the centre of the circle from the origin is 5.
We are talking of the max and min values of the absolute values of z not z itself, just to be clear. We cannot talk about maximum or minimum values for complex numbers themselves - it doesn't make sense. So it's the max and min of |z| we are looking for:
Get some graph paper or just make a sketch and draw the circle centred at (-3,4) with radius 2 and you will see that the circle does not cut through the x and y axes and does not enclose the origin.
Draw a line from the origin through the centre of the circle and extend it until it cuts the circle again. Choose a few points on the circle and draw a line from these various points on the circle to the origin. The lengths of these lines are the values of |z| at these points.
You will soon see that if you start from the point on the circle where the line through the centre intersects it (the one nearest the origin) then as you move round the circle from this point in a clockwise or anticlockwise direction, the distance from the point where you are to the origin always increases until you reach the point on the other side of the circle where the line through the centre intersects it. At this point if you continue to move round the circle then the distance to the origin starts to decrease again until you reach the point where you started.
It is clear that the 2 points of intersection with the circle, of the line through the origin and the centre of the circle,are the maximum and minimum distances of points on the circle from the origin.
Since we know that the circle has a radius of 2 then then the minimum distance from the origin of a point on the circle must be the difference between the distance of the centre from the origin and the radius, and the maximum distance from the origin of a point on the circle must be the sum of the distance of the centre from the origin and the radius.
This is very difficult to explain in words - it would be much easier if I were talking to you and had a diagram in my hand.
Thanks for pointing out that the maximum and minimum positions for z are missing from my published solution. I will amend my solution and re-publish it. Here is how I would calculate them:
Since we know that cos[arg(c)]=-3/5 and sin[arg(c)] is 4/5, and since arg(z_min)=arg(z_max)=arg(c), then
z_min=|z_min| {cos[arg(c)] + i sin[arg(c)]} and z_max=|z_max| {cos[arg(c)] + i sin[arg(c)]} and so
z_min=3 x -3/5 + i 3 x 4/5=-9/5 + i 12/5 and z_max=7 x -3/5 + i 7 x 4/5=-21/5 + i 28/5
I hope this helps you. If you have any more questions please ask. I would ideally like this forum to be used more than it is. I publish all my solutions but get very little feedback, except from you and Gary!
Thanks
Vasco
Nicolas,
ReplyDeleteIf you don’t mind someone else chiming in, would it help to know that the absolute value of c is the length of the segment from the origin of the coordinates to the centre c of the circle? The absolute value (length) |z| is measured on a line segment from the origin of the coordinates to the point z, which in this case is on the circle. It is not measured vertically unless for z = x + iy, x happens to be zero.
Gary
Gary: After writing my question I had noticed that Vasco had drawn a line from the origin to the centre and where it cuts the circle I got the values 3 and 7 by some rather laborious and roundabout trigonometry. But you have clarified for me what should have been blindingly obvious from the start and showed my calculations to be unnecessary!
DeleteHowever I do notice that Needham asks not only for the max and min values but also the position of z but which I presume he means the co-ordinates and therefore one does need to do the trigonometry?
In real number calculus one would get at min and max values by differentiating. Can one do the same in complex analysis?
I think my problem with Needham is that he goes too fast. I am used to more traditional text books where there are two or three examples of how to solve a particular problem followed by perhaps twenty or so exercise questions which serve to really drum the ideas into one.
Although the positions can be calculated as you suggest, I don't think this is in the spirit of what Needham is trying to get across at this point. I think he wants the reader to realise that the argument of the complex numbers c, z_min and z_max is the same and therefore since we know the sin and cosine of this argument and have previously calculated |z_min| and |z_max|, we can easily calculate z_min and z_max using z=r(cos(\theta)+isin(\theta) where theta is the argument referred to above.
DeleteThis is a replacement for July 9, 9:25.
DeleteIt’s easy to miss the forest for the trees. Some easy geometry with similar triangles will get those coordinates. Yes, the position would refer to the coordinates as Vasco writes them, e.g. (-3, 4). I get (-9/5, 12/5) for the position of z with min |z|. Of course you could also write z = -9/5 + 12i/5. Note that the diagram for part (ii) is a bit different from the general one that Vasco shows because the centre of the circle is left of the y coordinate.
As for complex derivatives, it looks like chapters 4 and 5 will get into that.
Needham assumes a lot of background and moves fast. I have gotten stuck at several points. On the other hand, the book forces review and integration of many topics and the polar notation z = re^(i theta) provides a way of calculating and thinking about positions in two dimensions that is often more efficient and intuitive than vector notation. I even finding myself applying it to ideas that I think of as coming from linear algebra, as when I calculated the minimum position above as (|c|-r) (c/|c|), where (c/|c|) is analogous to a unit vector. It’s a number of length 1 with the same angle as c. But Vasco provides a more instructive equation.
Oops. I wish we could edit. That z minimum should have been (-9/5, 12/5) or z = -9/5 + 12/5. -- Gary
ReplyDeleteGary
DeleteYou can always use cut and paste and then delete your original post and then resubmit.
Hi Nicolas and Gary
ReplyDeleteI have posted an answer to Nicolas' original question above immediately after his question.
Vasco
I have uploaded an amended version of exercise 6 chapter 1 which includes calculation of z_min and z_max.
ReplyDeleteVery neat and instructive.
DeleteGary
Chapter 1 Exercise 9: Would another solution be:
ReplyDeletecos x + isin x + cos(x+90) + isin(x+90) + cos(x+180) + isin(x+180) + cos(x+270) + isin(x+270) = 0 ?
What does x represent?
DeleteNicolas, I tried to improve the reasoning and correct a couple of errors in the description for problem 9 in Chapter 1.
ReplyDelete9. Let A, B, C, D be four points on the unit circle. If A + B + C + D = 0, show that the points must form a rectangle.
We assume that the points are described with complex numbers on the Argand plane and that the circle is drawn with center at the origin. Since A, B, C, and D could be any four points, we rotate the points equally (or we rotate the axes) until the x axis is parallel to side AB. We use A = a1 + a2 i, where a1 = Re[A] = cos Arg(A) and a2 = Im[A] = sin Arg(A), etc. to represent the points.
Diagram at https://www.dropbox.com/s/fd9e1c7ia0a2chp/ch1.ex9.4pts.on.circle.jpg
If the points sum to zero, they must still sum to zero after rotation, because rotation on the circle preserves the shape. Both the real and imaginary points must sum to zero. Then we can write:
We write:
A = a1 + a2 i
B = b1 + b2 i = -a1 + a2 i
(because the real points of A and B—being on the same vertical line—must cancel and the imaginary points must be equal)
C = c1 + c2 i = a1 - a2 i
(because, by the diagram, the imaginary points of A and C must cancel and the real points must be equal)
D = d1 + d2 i = -a1 - a2 i
(because, by the diagram, both the real points and the imaginary points of A and D must cancel).
Summing these up, they add up to zero. If they do not sum to zero, the four points can not describe a rectangle.
***************
It seems that every time I try to delete and add a comment, the original comment gets restored. This is my last attempt for this comment.
Gary
I don't think there are any errors in the description of exercise 9 chapter 1 in the book. Why do you think there are?
DeleteGary
Delete1. We don't have to assume that the circle is centred at the origin. When we say in maths THE unit circle, we mean a circle of unit radius centred at the origin. When we say A unit circle we mean a circle of unit radius without specifying the centre.
3. You are assuming that the points ABCD are the vertices of a rectangle - you have to prove this.
Obviously 3. above should be 2.
DeleteI think I was not seeing the continuation link at the bottom of the screen. I will try to clean up the redundant comments at the first opportunity. The PC and Google browser that I am using this week does not always display the delete link. In the meantime, please feel free to delete any redundant comments.
ReplyDeleteGary
As Gary has pointed out, there seems to be an error in the process for deleting posts/comments.
ReplyDeleteHi Gary
ReplyDeleteI have replied to one of your posts above that I think you wanted to delete. I can delete it for you, but I wanted to check with you first. Is it the one on 14th July at 4:48pm?