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### Use De Moivre's Theorem to show that if z = cos(q)+isin(q), then (z^n)+(z^-n) = 2cos(nq) and (z^n)-(z^-n)=2isin(nq).

De Moivre's Theorem states that if z = cos(q)+isin(q), then

zn = (cos(q)+isin(q))n = cos(nq)+isin(nq)

But then

z-n = cos(-nq)+isin(-nq).

Now, cos(-p)=cos(p), as cosine is a symmetric (even) function, and sin(-p)=-sin(p), as sine is an anti-symmetric (odd) fuction. Thus,

z-n  = cos(nq)-isin(nq).

The rest is just algebra:

zn+z-n = [cos(nq)+isin(nq)]+[cos(nq)-isin(nq)] = 2cos(nq).

zn-z-n = [cos(nq)+isin(nq)]-[cos(nq)-isin(nq)] = 2isin(nq).

2 years ago

Answered by Dorian, an A Level Further Mathematics tutor with MyTutor

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