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

Proof of De Moivre's Theorem

How is de Moivre’s Theorem proved?

• When written in Euler’s form the proof of de Moivre’s theorem is easy to see:

  • Using the index law of brackets: 

• However Euler’s form cannot be used to prove de Moivre’s Theorem when it is in modulus-argument (polar) form

• Proof by induction can be used to prove de Moivre’s Theorem for positive integers:

  • To prove de Moivre’s Theorem for all positive integers, n

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• STEP 1: Prove it is true for n = 1

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

• STEP 2: Assume it is true for n = k

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• STEP 3: Show it is true for n = k + 1

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  • According to the assumption this is equal to

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  • Using laws of indices and multiplying out the brackets:

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  • Letting i² = -1 and collecting the real and imaginary parts gives:

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  • Recognising that the real part is equivalent to cos(kθ  + θ ) and the imaginary part is equivalent to sin(kθ  + θ ) gives

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  • So de Moivre’s Theorem is true for n = k + 1

• STEP 4: Write a conclusion to complete the proof

  • The statement is true for n = 1, and if it is true for n = k it is also true for n = k + 1

  • Therefore, by the principle of mathematical induction, the result is true for all positive integers, n

• De Moivre’s Theorem works for all real values of n

  • However you could only be asked to prove it is true for positive integers