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yoda5412 052008, 03:06 AM 1. The problem statement, all variables and given/known data
Hello! :smile:
a)Prove that (a+bi)^n and (abi)^n, n \in \mathbb{N} are conjugate complex numbers;
b)Prove that quotient of any two numbers from the set of \sqrt[n]{1} is again number from the set of \sqrt[n]{1}
c)Prove that reciproca value of any number from the set of \sqrt[n]{1}, is again number from the set \sqrt[n]{1}
2. Relevant equations
z=r(cos\alpha+isin\alpha)
\bar{z}=r(cos\alphaisin\alpha)
w_k=\sqrt[n]{r}(cos\frac{\alpha + 2k\pi}{n}+isin{\alpha + 2k\pi}{n}) ; k=0,1,2,...,n1
3. The attempt at a solution
a) (a+bi)=r(cos\alpha+isin\alpha)
(a+bi)^n=r^n(cosn\alpha+isinn\alpha)
(abi)=r(cos\alphasin\alpha)
(abi)^n=r^n(cosn\alphasinn\alpha)
Is this enough to prove that they are conjugate complex numbers?
b)E_k=\sqrt[n]{1}=cos\frac{2k\pi}{n}+isin\frac{2k\pi}{n} , k=0,1,2,3,...,n1
Should I make like this?
E_n__1=\sqrt[n]{1}=cos\frac{2(n1)\pi}{n}+isin\frac{2(n1)\pi}{n}
What should I do next?
c)If I know how to prove b) I will prove c)
In this case just (\sqrt[n]{1})^^1=\sqrt[n]{1}, right?
BenTheMan 052008, 09:15 AM Hmmm...
It may be helpful to express your complex number as
(a+ib) = r e^{i\alpha}
where you can determine r and alpha in terms of a and b. This will help for b and c.
For a) I think you've got it.
BenTheMan 052008, 09:16 AM Except you forgot an i in your solution to a).
yoda5412 052008, 09:34 AM Yes, but iin b) I have not learnt to express the complex number like that... I should proof like the E_k equation.
QuarkHead 052008, 10:18 AM I think Ben is right. For (a), set (a+ib)=re^{i \alpha}. Then for (a+ib)^2 you will have re^{i \alpha}re^{i \alpha} = re^{2(i \alpha)}. Then having done that for arbitrary n, remember that, if (aib) = re^{i \alpha},\; re^{i \alpha}re^{i \alpha} = e^0 = 1, you have your proof!
For (b) I think it will suffice to show that there are ? roots to ^n\sqrt1 when n is even, and ? different ones when n is odd. For proof, assume otherwise, that is they are not different, for some arbitrary n. You should arrive at a contradiction. Then the quotient condition follows instantly.
As you say, (c) is immediate, if you can get the above
yoda5412 052008, 10:27 AM But I have never learnt about re^i^\alpha and this lection that I am learning is not connected with it..
QuarkHead 052008, 11:58 AM yoda: OK, I was being a bit sloppy  either you have missed something, or you are not being taught very thoroughly. Let me give a quick tutorial on this.
I will call a set an algebraic set if there is at least one operation that combines any and all elements of the set.
I will call an algebraic set closed if any legitimate operation on its elements results in another element of the same set.
So, obviously, \mathbb{R} is an algebraic set, in this sense. Here, the allowed operations are plus, minus, times, divide, power, square root etc. But, by the above, \mathbb{R} is not algebraically closed, since although 1 \in \mathbb{R},\; \sqrt{1} \;\notin \mathbb{R}, rather \sqrt{1} \in \mathbb{C}.
It is for this reason that mathematicians (bless them!) say that the real numbers are a subset of the complex numbers: \mathbb{R} \subset \mathbb{C}.
Now to your problem, or rather, any similar problem.
Since \mathbb{C} is algebraically closed, we may assume that any complex or real number can be " mapped" to any other complex number by a legitimate operation on either set. So, unless you deny that re^{i \alpha} is a number of some sort (this would be extremely unwise!), then you may assume there is some complex number z = x +iy = re^{i \alpha}.
Any better?
yoda5412 052008, 12:29 PM Yes, thanks. But the question is, is there any way to prove it through E_k?
QuarkHead 052008, 02:08 PM Yes, thanks. But the question is, is there any way to prove it through..........? I see. You are not asking for some deep understanding, rather you want help with your homework. Did you understand my last post? Did you understand Ben's?
Your "Yes thanks, but..." above is an insult to me and Ben, we both spent some time on our responses. Does that mean nothing to you?
It is obvious, at least to me, that you hadn't considered either of our posts seriously. That is an insult, especially as Ben and I showed how to think about your problem.
If you came here to learn some mathematics or physics, then ask away. If you came here for homework help, then I, for one, am not interested  figure it out for yourself.
yoda5412 052008, 02:29 PM First, sorry if I insult you in some way. I know that it took some of your value time. And no, I don't want to just solve the homework. I want to understand what is happening, that's why I am asking for help. But lets be real, we all know that for any n\sqrt[n]{1}, I will again get 1. But that's what is not the task asking for. The task is to prove that. And actually I don't understand your post so much. I need calculations, do you understand me? Thanks for your responses, and sorry again if I insulted you, it was not my intention.
AlphaNumeric 052008, 02:45 PM It is obvious, at least to me, that you hadn't considered either of our posts seriously. That is an insult, especially as Ben and I showed how to think about your problem.
If you came here to learn some mathematics or physics, then ask away. If you came here for homework help, then I, for one, am not interested  figure it out for yourself.It would seem Yoda is StMartin (I don't htink he's trying to hide this fact) and it would also seem that he makes next to no effort to do his homework.
Yes, I can understand getting stuck on a question here or there but he seems to be using this forum (and perhaps others) as a crutch. "Oh no, I cannot do this question after 30 seconds of thinking, off to Sciforums I go". I used to spend days on homework sheets. It wasn't uncommon that after 2 or 3 days of work I might have done 4~6 questions only. True, the questions I'm talking about were 'slightly' lengthier than the questions in this thread, but my point is that I spent ages thinking about it if I couldn't get it immediately.
Yoda, ask friends, discuss things with friends. A discussion with friends is good for two reasons. Firstly, you don't get given the answer blindly. Secondly, it's a back and fore with people who are doing the same stuff as you, learning the same stuff as you, who will have similar problems as you. Both parties gain something. Understanding 1 concept is better than parroting 10 answers.
Oh and 1^{\frac{1}{n}} is only 1 once. There's n1 other solutions which are not 1. You're essentially solving z^{n}=1 which is something you should have explained indetail in your notes. I teach and mark a 1st year course which covers precisely that. It's better you spend an hour reading through your notes and working through on a bit of paper any examples the lecturer uses than posting a thread here and waiting for replies.
I speak from experience, it's all too easy to read someone's answer and think "Oh of course, easy". It's always easy once someone else shows you. And as Professor Gowers (a Field Medalist) says, once someone has shown you the answer, you're forever robbed of the pleasure of solving it yourself.
yoda5412 052108, 12:48 AM First, I don't know who is StMartin. Secondly, you don't even know in what country I am living. English is not my first language, my professor don't know how to solve it, so obviously I am deseperate. While I am solving my other tasks, I am waiting somebody to help me here, I don't want just somebody to solve my problem. I need to understand it. So please, give me some calculations, or tell me I will do some calculations and I will know how to prove it. Thanks.
CptBork 052108, 01:57 AM I think question 1 is quite easy to do and can be done without resorting to Euler's identity.
First, to make sure we're using the same notation, I call \overline{j} the complex conjugate of j. Then in any introductory text to complex numbers, one of the most basic results they usually present is that for any complex numbers j and k, it is easy to show that \overline{jk}=\overline{j}\overline{k}. That is, the complex conjugate of a product of 2 numbers is equal to the product of the conjugates of the two numbers. This can be easily proved by writing j=a+bi,\ k=c+di and doing the calculation.
This can then be extended to products of three or more complex numbers by using induction. The final result is that \overline{j_1j_2j_3j_4j_5\ldots\j_m}=\overline{j_1 } \overline{j_2} \overline{j_3} \overline{j_4}\overline{j_5}\ldots\overline{j_m}
So in part a), because n is a natural number, you can then write \overline{(a+bi)^n}=(\overline{a+bi})^n and from here it's straightforward what you need to do to finish the problem.
In part b), suppose a^n=b^n=1. Then you can write (\frac{a}{b})^n=\frac{a^n}{b^n}=\frac{1}{1}=1, which shows that if a and b are nth roots of 1, then so is the quotient \frac{a}{b}. You can use similar reasoning to solve part c). Remember that there are always n solutions to the equation x^n=1. \ x=1 is always a solution, but there are n1 other solutions, including complex numbers (unless n=2, in which case x=1 is the only other solution).
Hope this helps!
I'm sorry... where do you get
k=c+di from? Don't you mean that the product of two conjugates is equal to the sum of the varaibles to the second power...??
(a+bi)(abi)=a^{2}b^{2}i^{2}
??? :)
AlphaNumeric 052108, 03:27 AM First, I don't know who is StMartinYet your opening post is identical in format to his opening posts here : http://www.sciforums.com/showthread.php?t=81146
CptBork 052108, 03:38 AM I'm sorry... where do you get
k=c+di from? Don't you mean that the product of two conjugates is equal to the sum of the varaibles to the second power...??
(a+bi)(abi)=a^{2}b^{2}i^{2}
??? :)
The theorem you're quoting is different from the one I'm referring to. Yours is referring to the product of a complex number with its own conjugate, i.e. z\overline{z}=(a+bi)(\overline{a+bi})=(a+bi)(abi)=a^2b^2i^2=a^2+b^2. Mine is referring to the conjugate of the product of any two complex numbers, i.e. \overline{z_1z_2}=\overline{(a+bi)(c+di)}. So what you're referring to is totally different.
I should have used z_1 instead of j and z_2 instead of k, that would have been more in line with conventional notation.
Well, no worries... my fault... by the way, its cool. I use
J=(a+bi)
J'=(abi)
to express them... so i knew what you where on about :)
yoda5412 052108, 04:31 AM I think question 1 is quite easy to do and can be done without resorting to Euler's identity.
First, to make sure we're using the same notation, I call \overline{j} the complex conjugate of j. Then in any introductory text to complex numbers, one of the most basic results they usually present is that for any complex numbers j and k, it is easy to show that \overline{jk}=\overline{j}\overline{k}. That is, the complex conjugate of a product of 2 numbers is equal to the product of the conjugates of the two numbers. This can be easily proved by writing j=a+bi,\ k=c+di and doing the calculation.
This can then be extended to products of three or more complex numbers by using induction. The final result is that \overline{j_1j_2j_3j_4j_5\ldots\j_m}=\overline{j_1 } \overline{j_2} \overline{j_3} \overline{j_4}\overline{j_5}\ldots\overline{j_m}
So in part a), because n is a natural number, you can then write \overline{(a+bi)^n}=(\overline{a+bi})^n and from here it's straightforward what you need to do to finish the problem.
In part b), suppose a^n=b^n=1. Then you can write (\frac{a}{b})^n=\frac{a^n}{b^n}=\frac{1}{1}=1, which shows that if a and b are nth roots of 1, then so is the quotient \frac{a}{b}. You can use similar reasoning to solve part c). Remember that there are always n solutions to the equation x^n=1. \ x=1 is always a solution, but there are n1 other solutions, including complex numbers (unless n=2, in which case x=1 is the only other solution).
Hope this helps!
CptBork, yes it helped. But is b) and c) valid for only real numbers in your post? Since we didn't present any complex numbers
Yet your opening post is identical in format to his opening posts here :
AlphaNumeric that is template from physicsforums. I am registered there, and also post same threads like I do here.
CptBork 052108, 03:36 PM CptBork, yes it helped. But is b) and c) valid for only real numbers in your post? Since we didn't present any complex numbers
AlphaNumeric that is template from physicsforums. I am registered there, and also post same threads like I do here.
b) and c) are valid for complex numbers the same as they are for real numbers. The algebra is exactly the same, and I didn't need to assume anything about whether or not a and b were real numbers in my derivation.
