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

# help solve (z+i)^n + (z-i)^n = 0 watch

1. I can't solve it. I ve tried dividing through but don't feel sure about equating with the nth root of unity b/c of the minus... Am I meant to take cases for if n iss odd and if it is even?...

I can't get even close to the answer the book provides :
cot[(2k+1)pi/2n}

2. Firstly, can you say about |z+i|/|z-i|? What does this mean in geometrical terms? So if z = x+iy, what does this tell us about y?
Secondly, what can you then say about arg(z+i), arg(z-i)? (Draw a sketch, using what you found out about z from the first bit). But (z+i)^n = - (z-i)^n gives you another result linking arg(z+i), arg(z-i). The result should then be fairly obvious.
3. |z+i|/|z-i| constitutes a scaling the moduli of z+i by the factor z-i. Should i.. .? |z+i|=-|z-i| ==> x^2 + y^2 = -1?? What does this tell me about y?

arg[(z+i)/(z-i)] = arg (z+i) - arg (z-i)...?

sorry for being stunted with this...
4. |z+i| = -|z-i| is nonsense: you know (or should know) that a modulus is always positive.

Let's break it down into small pieces.

You know that (z+i)^n = -(z-i)^n.
That is, (z+i)^n = (-1) (z-i)^n. Now take the modulus of both sides.
Now since modulus are always positive, you can take the nth root of both sides. What do you end up with?
5. so take
|z+i| = |-(z-i)|? and get y = 0 ?

and by taking the nth root (x+i) / (-x+i) = cos2kpi/n + i sin2kpi/n (call it a)

so solving for x = i (a-1) / (1+a) and use the subsititions a-1 = 2i(sinkpi/n)e^ikpi/n and a+1 = 2(coskpi/n)e^ikpi/n ??

even so all "i" obtain is x = -tankpi/n which is not the right (book's) answer
6. Since you're still here and DFranklin isn't I'll stick my nose in.

I think you established that z is real.

So z+i and z-i are opposite each other above and below the real axis.

When you raise them to the same power 'they' will rotate in opposite directions.

Where do they need to rotate to for their sum to be zero?

So what do the original angles need to be, before rotation?

So what must z be?

Do some sketches.

Sorry to you both if this is not a helpful post.
7. do they need to be perpendicular and do they have to then have opposite directions for their sum to be zero?

cot[(2k+1)pi/2n...?

p.s. thanks for taking the time...
8. (Original post by pb6883)
do they need to be perpendicular and do they have to then have opposite directions for their sum to be zero?
That doesn't make sense, for a start.
9. A silly question.

Is this equation meant to be true for all n?
10. yes..
11. (Original post by pb6883)
do they need to be perpendicular and do they have to then have opposite directions for their sum to be zero?
Did you mean that (z+i)^n and (z-i)^n lie on the imaginary axis?

If so you were correct.

So what's the arg of z+i in terms of n if arg(z+i)^n = pi/2 ?
12. (Original post by stevencarrwork)
A silly question.

Is this equation meant to be true for all n?
It's meant to be solved in terms of n.
13. (Original post by rnd)
Did you mean that (z+i)^n and (z-i)^n lie on the imaginary axis?

If so you were correct.

So what's the arg of z+i in terms of n if arg(z+i)^n = pi/2 ?
so arg(z+i) = pi/n2

and arg(z+i)/(z-i) = arg(z+i) - arg(z-i) = pi/n2 - arg(z-i) = kpi =>

arg(z-i) = kpi + pi/n2 => arctan(z-i) = arctan(-1/z) = arccot(z) = 2kpi + pi/n2 => z = cot(2k+1)pi/2n

i 'missing something i know, i 've rushed, plus is the geometric reasoning of them being on the imaginary axis enought (??) anyway i have to leave work now so i 'll look at it again when i get home,

14. I went for the approach of re-arranging to then writing which gave me an answer in terms of e, but I can't be bothered with the re-arranging needed to check I got the same answer as the one in your book

Spoiler:
Show

also checked with wolfram and seems to be correct
15. (Original post by matt2k8)
... I can't be bothered with the re-arranging needed to check I got the same answer as the one in your book
Unfortunately, the rearranging is most of the work. If this was a 10 mark question, I'd say your answer would be lucky to get 3 marks.

Don't get me wrong, for a lot of purposes, who cares as long as you get a usable answer. But GE and I were trying to lead the OP towards getting the answer in the form desired.
Spoiler:
Show
(z+i)^n + (z-i)^n = 0 <=> (z+i)^n = -(z-i)^n <=> {(z+i)/(z-i)}^n = -1 <=> (z+i)/(z-i) = (-1)^-n <=> z+i = z(-1)^-n - i(-1)^-n <=> z(1-(-1)^-n)=-i(1+(-1)^-n) <=> z=-i{(1+(-1)^-n) / (1-(-1)^-n)}

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: September 19, 2013
Today on TSR

Poll
Useful resources

### 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