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

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 $n\in {\mathrm{\mathbb{Z}}}^{+}$, i.e. positive integer values of $n$

  • in which case you can use proof by induction

• You need to prove that $\left[r \right(\cos \theta +\mathrm{isin}\theta ){]}^n=r^n(\cos n\theta +\mathrm{isin}n\theta )$ for $n\in {\mathrm{\mathbb{Z}}}^{+}$

• STEP 1
  Prove it is true for $n=1$

  • $\left[r \right(\cos \theta +\mathrm{isin}\theta ){]}^1=r^1(\cos 1\theta +\mathrm{isin}1\theta )= r(\cos \theta +\mathrm{isin}\theta )$

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

• STEP 2
  Assume it is true for $n=k$

  • $\left[r \right(\cos \theta +\mathrm{isin}\theta ){]}^k=r^k(\cos k\theta +\mathrm{isin}k\theta )$

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

  • $\left[r \right(\cos \theta +\mathrm{isin}\theta ){]}^{k+ 1}=\left(\right[r (\cos \theta +\mathrm{isin}\theta ){]}^k\left)\right(\left[r \right(\cos \theta +\mathrm{isin}\theta \left){]}^1\right)$

  • According to the assumption this is equal to

    • $\left(r^k\right(\cos k\theta +\mathrm{isin}k\theta \left)\right)\left(r \right(\cos \theta +\mathrm{isin}\theta \left)\right)$

  • Using laws of indices and multiplying out the brackets

    • $= r^{k+1}[\cos k\theta \cos \theta +\mathrm{icos}k\theta \sin \theta +\mathrm{isin}k\theta \cos \theta +{\mathrm{i}}^2\sin k\theta \sin \theta ]$

  • Using $i^2=-1$ and collecting the real and imaginary parts gives

    • $= r^{k+1}[\cos k\theta \cos \theta -\sin k\theta \sin \theta +\mathrm{i}(\cos k\theta \sin \theta +\sin k\theta \cos \theta )]$

  • Recognising that the real part is equivalent to $\cos \left(k\theta +\theta \right)$ and the imaginary part is equivalent to $\sin \left(k\theta +\theta \right)$ gives

    • $\left[r \right(\cos \theta +\mathrm{isin}\theta ){]}^k=r^{k+1}[\cos \left(k+1\right)\theta +\mathrm{isin}\left(k+1\right)\theta ]$

  • 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, $n\in {\mathrm{\mathbb{Z}}}^{+}$