You are Here: Home

# Solve (z-i)^n = (z+i)^n

Maths and statistics discussion, revision, exam and homework help.

Announcements Posted on
Find out how cards are replacing warnings on TSR...read more 03-12-2013
1. Re: Solve (z-i)^n = (z+i)^n
(Original post by 99wattr89)
..
Yes, has to be an integer. For example, suppose . Then . If you work out what real value of z gives arg(z+i) = pi / n, you'll find z = cot(pi / n).

[If you're wondering why the calculations imply a solution when theta is a multiple of pi: these are "psuedo-solutions" that you'd get if we pretended we could have z equalling + or - infinity.]
2. Re: Solve (z-i)^n = (z+i)^n
(Original post by DFranklin)
Yes, has to be an integer. For example, suppose . Then . If you work out what real value of z gives arg(z+i) = pi / n, you'll find z = cot(pi / n).

[If you're wondering why the calculations imply a solution when theta is a multiple of pi: these are "psuedo-solutions" that you'd get if we pretended we could have z equalling + or - infinity.]
Why is it OK to assume that it's equal to 1? Couldn't it be any integer?

Assuming that it is 1, giving arg(z+i) = Pi/n, I still don't know how to actually solve it. I feel like I've missed a big chunk of lectures or something (even though I haven't), because I've never done questions like this before and I don't see how to get z from the argument.

I have arctan (1/z) = Pi/n, but that doesn't give the answer. I don't understand this questions at all.
3. Re: Solve (z-i)^n = (z+i)^n
Can anyone help me make sense of this and prove the solution?
4. Re: Solve (z-i)^n = (z+i)^n
I'm still stuck here. Can anyone help explain this to me?
5. Re: Solve (z-i)^n = (z+i)^n
(Original post by 99wattr89)
Why is it OK to assume that it's equal to 1? Couldn't it be any integer?
Yes it could. I'm giving an example of how you should be calculating it, because trying to calculate it for an arbitrary integer seems to be throwing you off.

Assuming that it is 1, giving arg(z+i) = Pi/n, I still don't know how to actually solve it. I feel like I've missed a big chunk of lectures or something (even though I haven't), because I've never done questions like this before and I don't see how to get z from the argument.
Well, you know z is real. So the point (z+i) has co-ordinates (z, i) on an argand diagram. So you simply need to find what z has to be for the angle between (z, i) and the x-axis to be Pi/n. Draw a diagram, you get a right angled triangle.

Of course, you may have already done this, because you say...

I have arctan (1/z) = Pi/n, but that doesn't give the answer. I don't understand this questions at all.
If arctan(1/z) = pi/n. What does arccot(z) equal? So what does z equal?

Now, instead of supposing that arg(z+i) = Pi / n, suppose it equals k pi / n for some integer k. How does this change the reasoning above?
6. Re: Solve (z-i)^n = (z+i)^n
(Original post by DFranklin)
Yes it could. I'm giving an example of how you should be calculating it, because trying to calculate it for an arbitrary integer seems to be throwing you off.

Well, you know z is real. So the point (z+i) has co-ordinates (z, i) on an argand diagram. So you simply need to find what z has to be for the angle between (z, i) and the x-axis to be Pi/n. Draw a diagram, you get a right angled triangle.

Of course, you may have already done this, because you say...

If arctan(1/z) = pi/n. What does arccot(z) equal? So what does z equal?

Now, instead of supposing that arg(z+i) = Pi / n, suppose it equals k pi / n for some integer k. How does this change the reasoning above?
I think I see a bit more what's going on now. Thank you very much for all the help.

So, since cot gives the real value over the imaginary one, in the case that z is real, not complex, and the imaginary value is known to be 1, that means that cot gives z. I hadn't grasped this at all, but it makes sense now.

I think a lot of what was confusing me was that I thought n could vary in the solution, but that's not the case - n varies for different solutions, but the solution is for one unspecified value of n.

* Z has to be real, because of the relationship between z+i and z-i.

If you define z+i with argument Theta then because z-i is the reflection in the real axis, it has to have angle -Theta, which translates to 2Pi - Theta.

From the laws of arguments, that means that the argument of (z-i)n/(z+i)n is 2n(Pi - Theta) (because arguments follow the laws of logs). But from rearranging the initial equation you also get that this argument is equal to the argument of 1, which is given by 2mPi where m is an integer.

So setting those two results equal to eachother, you see that since the other terms all all integer multiples of 2Pi, 2nTheta has to be as well, for the equality to hold.

So you get nTheta/Pi = r (where r is an integer).

So Theta = rPi/n

So arg(z+i) = rPi/n

And cot Theta = Real/Imaginary Using * this is equal to z/Im = z/1 = z

So z = cot (rPi/n) where r is an integer.

So, I think all I need to do is justify why r would be from 1 to n-1. I'm not so sure about this, but it seems like it would be something to do with the imaginary component being 1 for all r/z/n, meaning that you can only get solutions that lie the region 0>Theta>Pi.

But I'm still trying to think of a way for that to translate into n-1 solutions. If you can give me any tips there it would be great.
7. Re: Solve (z-i)^n = (z+i)^n
(Original post by 99wattr89)
all I need to do is justify why r would be from 1 to n-1
Actually, r can be any integer (*). But , so there's only a certain number of possible solutions.

(*) It can't be a multiple of n, because then .
8. Re: Solve (z-i)^n = (z+i)^n
(Original post by DFranklin)
Actually, r can be any integer (*). But , so there's only a certain number of possible solutions.

(*) It can't be a multiple of n, because then .
Shouldn't it be ?

Since n is constant, and it can be any integer multiple of Pi?

I can see logically how because cot is cyclic, you only get unique solutions in the range between 0 and Pi, but I can't see how this proves that the number of solutions is n-1.
Last edited by 99wattr89; 02-04-2012 at 13:18.
9. Re: Solve (z-i)^n = (z+i)^n
(Original post by 99wattr89)
Shouldn't it be ?
No.

I can see logically how because cot is cyclic, you only get unique solutions in the range between 0 and Pi, but I can't see how this proves that the number of solutions is n-1.
Because your solutions are the values of cot, not the angle.

e.g. cot (pi / n) and cot ((n+1)pi / n) both give the same value for z.
10. Re: Solve (z-i)^n = (z+i)^n
(Original post by DFranklin)
No.

Because your solutions are the values of cot, not the angle.

e.g. cot (pi / n) and cot ((n+1)pi / n) both give the same value for z.
I think I get that now, thank you.

For the final part of the question it says hence find the roots of 7z6 -35z4 +21z2 = 1

Can you give me any hints as to how the first two parts of the question allow that to be solved? I've been trying to find some sort of method using DeMoivre's formula, but I don't see how to apply it here, so maybe that's not the method the question means me to use?
11. Re: Solve (z-i)^n = (z+i)^n
Well, what's (z+i)^n - (z-i)^n? Compare with what you've been asked to solve, decide what value of n will make them correspond, etc.
12. Re: Solve (z-i)^n = (z+i)^n
(Original post by DFranklin)
Well, what's (z+i)^n - (z-i)^n? Compare with what you've been asked to solve, decide what value of n will make them correspond, etc.
Rather counter-intuitively, n=6 is wrong but n=7 gives 2i times the target equation. I would never have found that in an exam.

So, given that the solutions for the equation in part 2 are known, does that just mean that the solutions for this new equation are the previous z values, since 2i is just a constant?

And the roots then do sum to zero because the z values cot gives are mirrored in the imaginary axis - cot gives zero at r3, and the lower and high r values cancel each other out.
13. Re: Solve (z-i)^n = (z+i)^n
Sounds right.

I'd say it's not a trivial exam question - I did have to spend about a minute thinking about where to start. [Worded like that it probably sounds really obnoxious/arrogant, but it does put it towards the harder side for these kinds of questions].
14. Re: Solve (z-i)^n = (z+i)^n
(Original post by DFranklin)
Sounds right.

I'd say it's not a trivial exam question - I did have to spend about a minute thinking about where to start. [Worded like that it probably sounds really obnoxious/arrogant, but it does put it towards the harder side for these kinds of questions].
Hah hah! I think after the amount of time you've spent helping me with this question you're entitled to feel pretty good about yourself!

Thanks for all your help and patience.

## Step 2: Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank

this is what you'll be called on TSR

2. this can't be left blank

never shared and never spammed

3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. By completing the slider below you agree to The Student Room's terms & conditions and site rules

2. Slide the button to the right to create your account

You don't slide that way? No problem.

Last updated: April 3, 2012
Study resources