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Degree: Mathematics (Bachelors) - Cambridge University
Second year Cambridge maths undergraduate looking to help A-Level Maths and physics students, and also STEP, AEA or MAT.
|Further Mathematics||A Level||£22 /hr|
|Maths||A Level||£22 /hr|
|Physics||A Level||£22 /hr|
|.STEP.||Uni Admissions Test||£25 /hr|
Nancy (Parent) June 9 2016
Nancy (Parent) June 10 2016
Nancy (Parent) June 14 2016
Nancy (Parent) April 13 2016
Let me rephrase this question slightly:
"Find all the roots of the equation x^n - 1 = 0."
We know by the fundamental theorem of algebra that an nth degree polynomial has *exactly* n roots. So the excersice has now been reduced to something as simple as: can you find n different numbers (call them x) such that x^n = 1.
Well we know 1 works. Let us call it eoiπ from now on. We still need to find the (n-1) other roots. The key to this is using the fact that 1 = eoiπ, e2iπ, e4iπ, ... So long as our number when raised by n goes to any one of these numbers, we are done. Well, we can see e2iπ/n does the job, and we can also see e4iπ/n does the job, so more generally e2riπ/n does the job for all positive integer r. Now we just need to find n of these numbers that are actually distinct (recall that there are infinitely many different ways of writing a number depending on how you write its argument, so while two numbers may be written differently they will actually be the same).
But fear not! If we look at e2riπ/n for 0 <= r < n, and r an integer, these are all distinct! (If they were not distinct then we could prove that eix = 1 for an x such that 0 < x < 2π, which is false).