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FP2 Complex Numbers/Functions

I think I'm ignoring something pretty obvious here.

I know that if I have something such as: z7=1 z^7 = 1 , then it will have multiple roots - 7 in the range π<θπ -\pi < \theta \leq \pi .

To work this out I'll have to convert the RHS into a different form using De Moivre, and then find the seventh root by (z7)17=RHS17 (z^7)^\frac{1}{7}=RHS^\frac{1}{7}. My issue with this is, doesn't this mean that I have a one to many function as raising both sides to the power 1/7 gives me 7 different values, and clearly this can't exist?
(edited 9 years ago)
Reply 1
Original post by ThatPerson
I think I'm ignoring something pretty obvious here.

I know that if I have something such as: z7=1 z^7 = 1 , then it will have multiple roots - 7 in the range π<θπ -\pi < \theta \leq \pi .

To work this out I'll have to convert the RHS into a different form using De Moivre, and then find the seventh root by (z7)(1/7)=RHS17 (z^7)^(1/7)=RHS^\frac{1}{7}. My issue with this is, doesn't this mean that I have a one to many function as raising both sides to the power 1/7 gives me 7 different values, and clearly this can't exist?



I'm not sure if I'm 100% correct here but;

Solving the equation using De Moivre's essentially only gives you one general expression for a root with the argument in terms of k, and you find each individual root by choosing suitable integer values of k that make the argument lie in π<θπ -\pi < \theta \leq \pi .

So what you're getting is not one to many since all the values are not the same! Just because it's the 7th root doesn't mean that all seven roots are equal. Hope I helped! :smile:

EDIT: oops I seem to have misunderstood your question!
(edited 9 years ago)
Reply 2
In the complex plane you cannot just raise both sides to a power like you can normally (unless the power is an itneger). The reason for this is exactly what you said, there are 7 different possible values. There are ways round this, but they are beyond what you need to know.
Reply 3
there is no problem with what you are doing...or maybe I am missing your point.
Over the reals x^2 =1 is exactly the same...
It will all make sense if you follow a mathematics degree, when you do a course of complex analysis.
Reply 4
Original post by Navo D.
I'm not sure if I'm 100% correct here but;

Solving the equation using De Moivre's essentially only gives you one general expression for a root with the argument in terms of k, and you find each individual root by choosing suitable integer values of k that make the argument lie in π<θπ -\pi < \theta \leq \pi .

So what you're getting is not one to many since all the values are not the same! Just because it's the 7th root doesn't mean that all seven roots are equal. Hope I helped! :smile:


Not sure I follow.

If they were all the same, then I'd have a one-to-one function. However as they are all different, I have a one-to-many function, where 1 is mapped to 7 different values.
Reply 5
Original post by james22
In the complex plane you cannot just raise both sides to a power like you can normally (unless the power is an itneger). The reason for this is exactly what you said, there are 7 different possible values. There are ways round this, but they are beyond what you need to know.


In the textbook, what I described above is essentially the method they use. Just to clarify, are you saying that this would be incorrect?
Reply 6
Original post by ThatPerson
In the textbook, what I described above is essentially the method they use. Just to clarify, are you saying that this would be incorrect?


It is technically incorrect, but it will work.
Reply 7
Original post by james22
It is technically incorrect, but it will work.


Ah I see. Thanks :smile:
Reply 8
Original post by ThatPerson
Not sure I follow.

If they were all the same, then I'd have a one-to-one function. However as they are all different, I have a one-to-many function, where 1 is mapped to 7 different values.


Yep I misunderstood your post, my bad! Maybe the rules/principles are different when considering the complex plane :eek: Because considering the reverse, we don't raise a single root to the nth power, instead just multiply the n roots out. Regardless, interesting observation - it seems rather counterintuitive!
Reply 9
Original post by ThatPerson
Ah I see. Thanks :smile:


Btw don't try using fractional powers with complex numbers in other circumstances. Things can go very wrong if you don't know how to correctly use them.
Reply 10
Consider x^2=1 over the reals!
There are two distinct roots for this equation, and in an analogous way the equation x^7 =1 has 7 distinct values.
The only difference is that on the first occasion the roots are both real but in the second you have 6 complex and 1 real.
If your bother is the way it is described " as raising both sides to the power of 1/7" please ignore as it will not aid your overall understanding at this stage.
Reply 11
Consider x^2=1 over the reals!
There are two distinct roots for this equation, and in an analogous way the equation x^7 =1 has 7 distinct values.
The only difference is that on the first occasion the roots are both real but in the second you have 6 complex and 1 real.
If your bother is the way it is described " as raising both sides to the power of 1/7" please ignore as it will not aid your overall understanding at this stage.

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