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2.5 Complex numbers

Syllabus
9231–2028–2029
Topic
2.5
Level
A2

Learning objectives

2.5.1de Moivre's theorem• understand de Moivre's theorem, for a positive or negative integer exponent, in terms of the geometrical effect of multiplication and division of complex numbers2.5.2de Moivre's theorem• prove de Moivre's theorem for a positive integer exponent e.g. by induction.2.5.3de Moivre's theorem• use de Moivre's theorem for a positive or negative rational exponent - to express trigonometrical ratios of multiple angles in terms of powers of trigonometrical ratios of the fundamental angle - to express powers of sin i and cos i in terms of multiple angles - in the summation of series - in finding and using the nth roots of unity. e.g. expressing cos5 i in terms of cos i or tan5 i in terms of tan i. e.g. expressing sin6 i in terms of cos2 i, cos4 i and cos6 i. e.g. using the 'C + iS ' method to sum series such as 1 sinn r r r n i = eo/.

Complex powers repeat a scale and a rotation

operation modulus argument geometric effect
z1z2z_1z_2 r1r2r_1r_2 θ1+θ2\theta_1+\theta_2 scale by r2r_2, rotate by θ2\theta_2
z1/z2z_1/z_2 r1/r2r_1/r_2 θ1−θ2\theta_1-\theta_2 divide the scale, undo the rotation
znz^n rnr^n nθn\theta repeat the scale and rotation nn times

Writing $\operatorname{cis}\theta=\cos\theta+i\sin\theta$, de Moivre's theorem for an integer $n$ is[r\operatorname{cis}\theta]^n=r^n\operatorname{cis}(n\theta).Arguments differing by $2\pi$ describe the same point.

For $n<0$, division explains the result. For example,(2\operatorname{cis}(\pi/6))^{-2}=\frac{1}{(2\operatorname{cis}(\pi/6))^2}=\frac14\operatorname{cis}(-\pi/3).Themodulusstayspositivewhileitsreciprocalpowershrinksthepointtowardtheorigin.The modulus stays positive while its reciprocal power shrinks the point toward the origin.

A power changes both modulus and argument. A negative exponent reverses the scale and rotation; it does not create a negative modulus, and it requires z≠0z\ne0.

One more multiplication is the induction step

For every positive integer $n$, the claim isP(n):\quad(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta).

Base case n=1n=1: both sides are cos⁡θ+isin⁡θ\cos\theta+i\sin\theta, so P(1)P(1) is true.

Assume $P(k)$ is true. Then\begin{aligned}(\cos\theta+i\sin\theta)^{k+1}&=\cos(k\theta)+i\sin(k\theta)\&=\cos((k+1)\theta)+i\sin((k+1)\theta),\end{aligned}because the real and imaginary parts are exactly the cosine and sine addition formulae. Thus $P(k)\Rightarrow P(k+1)$.

Therefore the theorem holds for every positive integer nn by mathematical induction. Multiplying by rnr^n gives [r(cos⁡θ+isin⁡θ)]n=rn(cos⁡nθ+isin⁡nθ)[r(\cos\theta+i\sin\theta)]^n=r^n(\cos n\theta+i\sin n\theta).

Several checked values are evidence for a conjecture, not a proof. State the induction hypothesis and show the k+1k+1 multiplication; do not replace this objective with the separate formula for nth roots.

Choose the de Moivre route that matches the target

For a multiple-angle identity, expand $(\cos\theta+i\sin\theta)^n$ and equate real or imaginary parts. For example,\cos5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta.Dividing matching real and imaginary expressions by a power of $\cos\theta$ can produce identities in $\tan\theta$.

To express powers in multiple angles, set $z=\cos\theta+i\sin\theta$. Since $z+z^{-1}=2\cos\theta$ and $z^m+z^{-m}=2\cos m\theta$, expanding $(z+z^{-1})^4$ gives\cos^4\theta=\frac18(\cos4\theta+4\cos2\theta+3).Use $z-z^{-1}=2i\sin\theta$ for sine powers.

Foratrigonometricseries,combineitasFor a trigonometric series, combine it asC+iS=\sum_{r=0}^{n}a_r(\cos r\theta+i\sin r\theta)=\sum_{r=0}^{n}a_rz^r,\qquad z=e^{i\theta}.Evaluate the resulting algebraic or geometric series, rationalise if needed, then take its real part for $C$ and imaginary part for $S$.

If $w^n=R\operatorname{cis}\phi$, all roots arew_k=R^{1/n}\operatorname{cis}\left(\frac{\phi+2\pi k}{n}\right),\qquad k=0,1,\ldots,n-1.For the nth roots of unity, $R=1$ and $\phi=0$, so the roots are equally spaced and their arguments differ by $2\pi/n$.

For zp/qz^{p/q} in lowest terms, solve wq=zpw^q=z^p and include all distinct branches; a negative pp first uses the reciprocal, so z≠0z\ne0. Dividing one principal argument by qq gives only one value, not the complete rational-power or root set.