De Moivre's theoremAQA A-Level Further Maths: Revision notes

De Moivre's theorem states that for any integer $n$, $(\cos\theta+\mathrm{i}\sin\theta)^n=\cos n\theta+\mathrm{i}\sin n\theta.$ In exponential form it is $\left(\mathrm{e}^{\mathrm{i}\theta}\right)^n=\mathrm{e}^{\mathrm{i}n\theta}$. For a general complex number, $\left[r(\cos\theta+\mathrm{i}\sin\theta)\right]^n=r^n(\cos n\theta+\mathrm{i}\sin n\theta)$: raise the modulus to the power, multiply the argument by $n$. Proof for positive integers is by induction: if it holds for $k$, then $(\cos\theta+\mathrm{i}\sin\theta)^{k+1}=(\cos k\theta+\mathrm{i}\sin k\theta)(\cos\theta+\mathrm{i}\sin\theta)=\cos(k+1)\theta+\mathrm{i}\sin(k+1)\theta$ using the addition formulae. Negative powers follow from $\frac{1}{\cos\theta+\mathrm{i}\sin\theta}=\cos\theta-\mathrm{i}\sin\theta$.

Convert to modulus-argument form, apply the theorem, then convert back if asked. Example: $w=1+\mathrm{i}=\sqrt2\left(\cos\frac{\pi}{4}+\mathrm{i}\sin\frac{\pi}{4}\right)$. Then $w^{10}=(\sqrt2)^{10}\left(\cos\frac{10\pi}{4}+\mathrm{i}\sin\frac{10\pi}{4}\right)=32\left(\cos\frac{5\pi}{2}+\mathrm{i}\sin\frac{5\pi}{2}\right)=32\mathrm{i}$. For a negative power, $z^{-n}=\cos(-n\theta)+\mathrm{i}\sin(-n\theta)=\cos n\theta-\mathrm{i}\sin n\theta$ when $|z|=1$.

To express $\cos n\theta$ and $\sin n\theta$ in terms of powers of $\cos\theta$ and $\sin\theta$: expand $(\cos\theta+\mathrm{i}\sin\theta)^n$ binomially, then equate real and imaginary parts with $\cos n\theta+\mathrm{i}\sin n\theta$. Write $c=\cos\theta$, $s=\sin\theta$. For $n=3$: $(c+\mathrm{i}s)^3=c^3+3\mathrm{i}c^2s-3cs^2-\mathrm{i}s^3$. Real part: $\cos3\theta=c^3-3cs^2=4c^3-3c$ (using $s^2=1-c^2$). Imaginary part: $\sin3\theta=3c^2s-s^3=3s-4s^3$. Use the identity to solve equations: $8x^3-6x-1=0$ with $x=\cos\theta$ becomes $2\cos3\theta-1=0$, so $\cos3\theta=\frac12$.

Let $z=\cos\theta+\mathrm{i}\sin\theta$. Then $z^n+\frac{1}{z^n}=2\cos n\theta$ and $z^n-\frac{1}{z^n}=2\mathrm{i}\sin n\theta$. Example: $(2\cos\theta)^4=\left(z+\frac1z\right)^4=z^4+4z^2+6+\frac{4}{z^2}+\frac{1}{z^4}=2\cos4\theta+8\cos2\theta+6$, so $\cos^4\theta=\frac18(\cos4\theta+4\cos2\theta+3)$.

To sum a series such as $\sum r^k\cos k\theta$, treat it as the real part of $\sum\left(r(\cos\theta+\mathrm{i}\sin\theta)\right)^k$, which is geometric (by de Moivre). The sum to infinity of a geometric series with first term $a$ and ratio $\rho$ is $\frac{a}{1-\rho}$ provided $|\rho|<1$. Example: $C+\mathrm{i}S=\sum_{k=0}^{\infty}\left(\frac{\cos\theta+\mathrm{i}\sin\theta}{2}\right)^k=\frac{2}{2-\cos\theta-\mathrm{i}\sin\theta}$. Multiply by the conjugate of the denominator: $\frac{2(2-\cos\theta+\mathrm{i}\sin\theta)}{5-4\cos\theta}$. So $C=\frac{4-2\cos\theta}{5-4\cos\theta}$ and $S=\frac{2\sin\theta}{5-4\cos\theta}$.