In one line
De Moivre's theorem says \((\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta\): raising this number to the power \(n\) simply multiplies the angle by \(n\).
Explained simply
Imagine a clock hand of length 1 pointing at angle \(\theta\). Multiplying two such numbers adds their angles, so multiplying the same one by itself \(n\) times just turns the hand \(n\) times as far. The length stays 1 because \(1^n=1\).
Statement: for every integer \(n\) (positive, negative or zero), \((\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta\). If \(n=p/q\) is a rational number, \(\cos n\theta+i\sin n\theta\) is one of the \(q\) values of \((\cos\theta+i\sin\theta)^{p/q}\). In exponential form it is just \((e^{i\theta})^n=e^{in\theta}\), which is the ordinary law of indices.
Why it matters: without it, \((\cos\theta+i\sin\theta)^{10}\) would need ten multiplications. With it, the answer is one line. It is the engine behind every topic in this module: expansions of \(\cos n\theta\), powers of \(\cos\theta\), powers of any complex number and finding roots.
How it works
- Write each factor in the form \(\cos\alpha\pm i\sin\alpha\). Note \(\cos\theta-i\sin\theta=\cos(-\theta)+i\sin(-\theta)=e^{-i\theta}\).
- Convert sines written first: \(\sin\theta+i\cos\theta=i(\cos\theta-i\sin\theta)=i\,e^{-i\theta}\).
- Replace each bracket by \(e^{i(\text{angle})}\), apply powers, then add or subtract exponents.
- Convert back using \(e^{i\phi}=\cos\phi+i\sin\phi\).
Key formula
\((\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta\), \(\ (\cos\theta-i\sin\theta)^n=\cos n\theta-i\sin n\theta\).
\(\theta\) is any real angle in radians, \(n\) an integer. For a general complex number \(z=r(\cos\theta+i\sin\theta)\): \(z^n=r^n(\cos n\theta+i\sin n\theta)\).
Quick example
Simplify \(\dfrac{(\cos2\theta+i\sin2\theta)^3}{(\cos\theta-i\sin\theta)^2}\).
Numerator \(=(e^{2i\theta})^3=e^{6i\theta}\). Denominator \(=(e^{-i\theta})^2=e^{-2i\theta}\). Quotient \(=e^{6i\theta-(-2i\theta)}=e^{8i\theta}=\cos8\theta+i\sin8\theta\).
Common mistakes
- Treating \(\cos\theta-i\sin\theta\) as \(e^{i\theta}\); it is \(e^{-i\theta}\).
- Applying the theorem to \(\sin\theta+i\cos\theta\) directly as if it were \(\cos\theta+i\sin\theta\).
- Forgetting \(r^n\) when the modulus is not 1, e.g. \((1+i)^4\) is \((\sqrt2)^4(\cos\pi+i\sin\pi)=-4\), not \(-1\).
How MU asks it
The syllabus asks only for the statement (no proof), so expect a 3-mark Q.1 part like "Simplify \(\dfrac{(\cos3\theta+i\sin3\theta)^5(\cos\theta-i\sin\theta)^3}{(\cos5\theta+i\sin5\theta)^7}\)". Write the exponential form of each bracket, show the exponent arithmetic in one line, and give the final answer as \(\cos k\theta+i\sin k\theta\). The theorem is then used in every expansion and roots question of this module.