- #1
theriel
- 27
- 0
Root of "n" degree of the number "n"
Hello!
Frankly speaking I am not sure whether this is proper forum, because I am not sure if this exercise involves calculus or not...
The task is:
Consider the numbers "root of "n" of n degree", n >= 2. How many of them are equal? Justify your answer.
I wanted to use complex numbers:
[itex]
n=re^{\mathit{i*A}}
[/itex]
[itex]
w_{k}=\sqrt[{n}]{r} * e^{(i * \frac{(2kpi + A)}{n})}
[/itex]
[itex]
w_{k+1}=w_{k}\ast e^{(i\ast \frac{(2pi)}{n})}
[/itex]
But frankly speaking... I do not know that to do with that #-/. I know that for any rational "n", the numbers "after some time" will be the same (the angle A will be over 360 and hence repeat)... Is that good way to follow?
Thank you for suggestions,
Theriel
Hello!
Frankly speaking I am not sure whether this is proper forum, because I am not sure if this exercise involves calculus or not...
The task is:
Consider the numbers "root of "n" of n degree", n >= 2. How many of them are equal? Justify your answer.
I wanted to use complex numbers:
[itex]
n=re^{\mathit{i*A}}
[/itex]
[itex]
w_{k}=\sqrt[{n}]{r} * e^{(i * \frac{(2kpi + A)}{n})}
[/itex]
[itex]
w_{k+1}=w_{k}\ast e^{(i\ast \frac{(2pi)}{n})}
[/itex]
But frankly speaking... I do not know that to do with that #-/. I know that for any rational "n", the numbers "after some time" will be the same (the angle A will be over 360 and hence repeat)... Is that good way to follow?
Thank you for suggestions,
Theriel
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