Edison R.

Unavailable

Mathematics (Bachelors) - Cambridge University

4.7

This tutor is also part of our Schools Programme. They are trusted by teachers to deliver high-quality 1:1 tuition that complements the school curriculum.

21 completed lessons

Second year Cambridge maths undergraduate looking to help A-Level Maths and physics students, and also STEP, AEA or MAT.

Second year Cambridge maths undergraduate looking to help A-Level Maths and physics students, and also STEP, AEA or MAT.

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Ratings & Reviews

4.7from 20 customer reviews

Nancy (Parent from Solihull)

June 9 2016

Edi is very knowledgeable in Maths and Further maths! Highly recommend him!

Nancy (Parent from Solihull)

June 10 2016

An excellent tutor, great help! Thank you so much!

Nancy (Parent from Solihull)

June 14 2016

Nancy (Parent from Solihull)

April 13 2016

An excellent tutor! Thank you!

Qualifications

MathematicsA-level (A2)A*
Further MathematicsA-level (A2)A*
PhysicsA-level (A2)A*

Subjects offered

SubjectQualificationPrices
Further MathematicsA Level£22 /hr
MathsA Level£22 /hr
PhysicsA Level£22 /hr

Find the nth roots of unity.

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, e2, e4, ... 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 e4/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).

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, e2, e4, ... 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 e4/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).

3 years ago

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