(z+i)^{5} + (zi)^{5} = 0
I've tried everything I can think of with this question, and I just keep getting that z = 0.
Can someone help me get started with the proper method?
Solve (z+i)^5 + (zi)^5 = 0
Announcements  Posted on  

TSR's new app is coming! Sign up here to try it first >>  17102016 

 Follow
 1
 10062012 19:23
Post rating:1 
 Follow
 2
 10062012 19:31
Have you tried expanding out the brackets and getting a polynomial in z?

 Follow
 3
 10062012 19:35
(Original post by ForGreatJustice)
Have you tried expanding out the brackets and getting a polynomial in z?Post rating:1 
 Follow
 4
 10062012 19:38
(Original post by 99wattr89)
(z+i)^{5} + (zi)^{5} = 0
I've tried everything I can think of with this question, and I just keep getting that z = 0.
Can someone help me get started with the proper method?
By considering the modulus of each side, show that z must be real.
Once you know z is real, suppose . By drawing a diagram, work out what must be.
Finally compare arguments of both sides to find what arg(z+i) must be and hence find z. 
 Follow
 5
 10062012 20:32
(Original post by DFranklin)
Rewrite as (z+i)^5 = (zi)^5.
By considering the modulus of each side, show that z must be real.
Once you know z is real, suppose . By drawing a diagram, work out what must be.
Finally compare arguments of both sides to find what arg(z+i) must be and hence find z.
Also when taking the moduli, should I get z+i = zi ? I can't tell if there should be a minus or not.
If there is a minus, then I've proven z is real.
From there I get that arg(zi) = 2Pi  Theta, and from that I get that Theta = 2Pi Theta, so that means Theta = Pi
So that means that (z+i) = re^{iPi(1+2n)}, so tan[Pi(1+2n)] = 1/(z+i), so z = cot[Pi(1+2n)]  i
I think all of that is correct, but I'm struggling to get the desired answer for z. It's meant to come out as z = cot[Pi(1+2n)/10], n = 0, 1, 2, 3, 4. 
 Follow
 6
 10062012 20:47
In this case, I knew z would be real because I've helped people do this question about 6 times on here. But it's not hard to show.
There should not be a minus sign. You have z+i = zi. Hint for proving z is real: For two complex numbers a and b, ab is the same as the distance between a and b.
I get that arg(zi) = 2Pi  Theta
Obviously your follow on working from this point is incorrect. 
 Follow
 7
 10062012 20:56
(Original post by DFranklin)
I really wouldn't recommend this method.
Looking at the coefficients, you know all the odd powers of i cancel, so you're just left with
2z^5 20z^3 + 10z = 2z(z^4  10z^2 + 5) which is a quadratic in z^2 and really easy to solve? 
 Follow
 8
 10062012 21:07
(Original post by CHY872)
Just wondering, why not?
Looking at the coefficients, you know all the odd powers of i cancel, so you're just left with
2z^5 20z^3 + 10z = 2z(z^4  10z^2 + 5) which is a quadratic in z^2 and really easy to solve? 
 Follow
 9
 10062012 22:46
(Original post by DFranklin)
In this case, I knew z would be real because I've helped people do this question about 6 times on here. But it's not hard to show.
There should not be a minus sign. You have z+i = zi. Hint for proving z is real: For two complex numbers a and b, ab is the same as the distance between a and b.
Depends on how you define arg, but it's more likely you want arg(zi) = theta. But I don't know why you think this implies "Theta = 2Pi Theta"; it doesn't.
Obviously your follow on working from this point is incorrect.
So, I have that arg(z+i) = Theta = arg(zi)
Also, (z+i)^{5}/(zi)^{5} = 1 means that arg[(z+i)^{5}/(zi)^{5}] = Pi (but strictly speaking it's Pi(1+2n), right? Since there are going to be multiple solutions.)
So, that means that (z+i)/(zi) = 1, right? Since the 5th root of 1 is 1.
So then arg(a/b) = arg(a)  arg(b) means that 2 Theta = Pi so Theta = 45 degrees, so z = 1.
Unfortunately that's totally wrong. But I can't work out what I'm doing wrong. I just don't know what it is I should do. 
 Follow
 10
 11062012 01:08
(Original post by 99wattr89)
...
As for the other parts, you seem to be skipping an important step and this leads you nowhere.
After you arrive at the expression
you could simplify your life a bit by transforming it into
, where .
Spoiler:Show
Hope this helps.Last edited by jack.hadamard; 11062012 at 12:00. Reason: Late night Maths. 
 Follow
 11
 11062012 11:40
(Original post by jack.hadamard)
The explanation for the first part is acceptable, though you may be required for more details.
As for the other parts, you seem to be skipping an important step and this leads you nowhere.
After you arrive at the expression
you could simplify your life a bit by transforming it into
, where .
Spoiler:Show
Hope this helps.
I have Theta = Pi(1+2n)/10
So arg(z+i) = Pi(1+2n)/10
The thing is that from there I get that tan Theta = i/z
Giving z = icot(Theta)
There shouldn't be an i though, z is real. 
 Follow
 12
 11062012 12:04
(Original post by 99wattr89)
That definitely does help, thank you.
I have Theta = Pi(1+2n)/10
So arg(z+i) = Pi(1+2n)/10
The thing is that from there I get that tan Theta = i/z
Giving z = icot(Theta)
There shouldn't be an i though, z is real.
Use your diagram again for the second part.
To deduce the tangent, you used a triangle with a side of length ? 
 Follow
 13
 11062012 14:32
(Original post by jack.hadamard)
I hope you noticed the hint with . If you are to find the fifth roots of , you do not take the fifth root of both sides first.
Use your diagram again for the second part.
To deduce the tangent, you used a triangle with a side of length ?
I don't know what else to do with w, since I already used it in my last attempt. I also don't understand what you mean when you say that you don't take the 5th root of both sides.
I did use a triangle with base z and height i.Last edited by 99wattr89; 11062012 at 14:42. 
 Follow
 14
 11062012 14:57
(Original post by 99wattr89)
I'm sorry, I don't understand.
I did use a triangle with base z and height i.
Spoiler:Show
The basic geometry of the complex plane, in ordinary terms, is no different from simple trigonometry.
Distances between complex numbers are real, and what you have said in your statement is that which is clearly false.
Draw the triangle again, the base is . What is its height again?Post rating:2 
 Follow
 15
 11062012 16:40
(Original post by jack.hadamard)
That's fine, I will attempt to explain it otherwise.
Spoiler:Show
The basic geometry of the complex plane, in ordinary terms, is no different from simple trigonometry.
Distances between complex numbers are real, and what you have said in your statement is that which is clearly false.
Draw the triangle again, the base is . What is its height again?
I'm really sorry about that, that was terrible of me. Thank you for being magnificently patient, I have the proper answer now.
[ z = 1/tan(Pi/10 + 2nPi/10) = cot(Pi(1+2n)/10) ]
Write a reply…
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following:
Sign in
Not got an account? Sign up now
Updated: June 11, 2012
Share this discussion:
Tweet
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.