If z=r(costheta+isintheta), then zn=rn(cos ntheta+isin ntheta) for positive integers n. For z 0, the formula also holds for zero and negative integer powers. Example (2 e(ipi)/(3))4=16 e(4pi i)/(3)=-8-8 3,i. Proof for positive integers Let P(n) be (costheta+isintheta)n=cos ntheta+isin ntheta. The case n=1 is immediate. Assume P(k) for some positive integer k. Using the polar multiplication rule, (costheta+isintheta)k+1=(cos ktheta+isin ktheta)(costheta+isintheta) =cos((k+1)theta)+isin((k+1)theta). Thus P(k+1) follows from P(k). By induction, the result holds for all positive integers n; multiplying by rn gives the general form. Rational powers require all possible arguments before dividing an angle. The roots method in the next section supplies all values; taking only a principal angle can miss roots.

If \(z=r(\cos\theta+i\sin\theta)\), then \(z^n=r^n(\cos n\theta+i\sin n\theta)\) for positive integers \(n\). For \(z\ne0\), the formula also holds for zero and negative integer powers.

\(\left(2\mathrm e^{\frac{i\pi}{3}}\right)^4=16\mathrm e^{\frac{4\pi i}{3}}=-8-8\sqrt3\,i\).

Let \(P(n)\) be \((\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta\). The case \(n=1\) is immediate.

Assume \(P(k)\) for some positive integer \(k\). Using the polar multiplication rule,

\(\begin{aligned}(\cos\theta+i\sin\theta)^{k+1}&=(\cos k\theta+i\sin k\theta)(\cos\theta+i\sin\theta)\\&=\cos((k+1)\theta)+i\sin((k+1)\theta).\end{aligned}\)

Thus \(P(k+1)\) follows from \(P(k)\). By induction, the result holds for all positive integers \(n\); multiplying by \(r^n\) gives the general form.

Rational powers require all possible arguments before dividing an angle. The roots method in the next section supplies all values; taking only a principal angle can miss roots.

Complex Numbers

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

De Moivre’s theorem

De Moivre’s theorem