Proof of De Moivre's Theorem (DP IB Analysis & Approaches (AA)): Revision Note

Amber

Written by: Amber

Reviewed by: Mark Curtis

Updated on

Proof of De Moivre's theorem

How do I prove de Moivre’s theorem?

  • You only need to be able to prove de Moivre's theorem for n element of straight integer numbers to the power of plus, i.e. positive integer values of n

    • in which case you can use proof by induction

  • You need to prove that left square bracket r blank left parenthesis cos invisible function application theta plus isin invisible function application theta right parenthesis right square bracket to the power of n equals r to the power of n left parenthesis cos invisible function application n theta plus isin invisible function application n theta right parenthesis for n element of straight integer numbers to the power of plus

  • STEP 1
    Prove it is true for n equals 1

    • left square bracket r blank left parenthesis cos invisible function application theta plus isin invisible function application theta right parenthesis right square bracket to the power of 1 equals r to the power of 1 left parenthesis cos invisible function application 1 theta plus isin invisible function application 1 theta right parenthesis equals blank r left parenthesis cos invisible function application theta plus isin invisible function application theta right parenthesis blank

    • So de Moivre’s Theorem is true for n equals 1

  • STEP 2
    Assume it is true for n equals k

    • left square bracket r blank left parenthesis cos invisible function application theta plus isin invisible function application theta right parenthesis right square bracket to the power of k equals r to the power of k left parenthesis cos invisible function application k theta plus isin invisible function application k theta right parenthesis blank

  • STEP 3
    Show it is true for n equals k plus 1

    • left square bracket r blank left parenthesis cos invisible function application theta plus isin invisible function application theta right parenthesis right square bracket to the power of k plus blank 1 end exponent equals left parenthesis left square bracket r blank left parenthesis cos invisible function application theta plus isin invisible function application theta right parenthesis right square bracket to the power of k right parenthesis left parenthesis left square bracket r blank left parenthesis cos invisible function application theta plus isin invisible function application theta right parenthesis right square bracket to the power of 1 right parenthesis blank

    • According to the assumption this is equal to

      • left parenthesis r to the power of k left parenthesis cos invisible function application k theta plus isin invisible function application k theta right parenthesis right parenthesis left parenthesis r blank left parenthesis cos invisible function application theta plus isin invisible function application theta right parenthesis right parenthesis

    • Using laws of indices and multiplying out the brackets

      • equals blank r to the power of k plus 1 end exponent left square bracket cos invisible function application k theta cos invisible function application theta plus icos space k theta space sin invisible function application theta plus isin invisible function application k theta cos invisible function application theta plus straight i squared sin invisible function application k theta sin invisible function application theta right square bracket

    • Using i squared equals negative 1 and collecting the real and imaginary parts gives

      • equals blank r to the power of k plus 1 end exponent left square bracket cos invisible function application k theta cos invisible function application theta minus sin invisible function application k theta sin invisible function application theta plus straight i left parenthesis cos space k theta space sin invisible function application theta plus sin invisible function application k theta cos invisible function application theta right parenthesis right square bracket

    • Recognising that the real part is equivalent to cos open parentheses k theta plus theta close parentheses and the imaginary part is equivalent to sin open parentheses k theta plus theta close parentheses gives

      • left square bracket r blank left parenthesis cos invisible function application theta plus isin invisible function application theta right parenthesis right square bracket to the power of k equals r to the power of k plus 1 end exponent left square bracket cos invisible function application open parentheses k plus 1 close parentheses theta plus isin invisible function application invisible function application open parentheses k plus 1 close parentheses theta right square bracket blank

    • So de Moivre’s Theorem is true for n equals k plus 1

  • STEP 4
    Write a conclusion to complete the proof

    • The statement is true for n equals 1, and if it is true for n equals k it is also true for n equals k plus 1

    • Therefore, by the principle of mathematical induction, the result is true for all positive integers, n element of straight integer numbers to the power of plus

Examiner Tips and Tricks

De Moivre's theorem actually holds for all real values of n, i.e. n element of straight real numbers, but you will only be asked to prove it for positive integer values of n, i.e. n element of straight integer numbers to the power of plus.

Worked Example

If z equals r open parentheses cos theta plus isin theta close parentheses, prove that

z to the power of n equals r to the power of n left parenthesis cos invisible function application n theta plus isin invisible function application n theta right parenthesis

for all positive integers n.

1-9-3-ib-aa-hl-proof-of-de-moivres-theorem-we-solution

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Amber

Author: Amber

Expertise: Maths Content Creator

Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.

Mark Curtis

Reviewer: Mark Curtis

Expertise: Maths Content Creator

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.