de Moivre's formula

Complex numbers: The complex plane

De Moivre's formula

In the notation of imaginary $e$ -powers various trigonometric formulas can be derived in a simple manner.

$(\cos\varphi + \mathrm{i}\sin\varphi)^n=\cos(n\,\varphi) + \mathrm{i}\,\sin(n\,\varphi)\qquad\text{for any integer }n$

This is a consequence of the second equality $\left(\e^{z}\right)^n=\e^{n\cdot z}$ from the calculation rules for imaginary $e$ -powers and two applications of Euler's formula: $\begin{aligned}\left(\cos(\varphi)+\ii\sin(\varphi)\right)^n &=\left(\e^{\ii\,\varphi}\right)^n \\ &\phantom{abcdwxyz}\color{blue}{\text{Euler's formula}} \\ &= \e^{n\,\varphi\,\ii} \\ &\phantom{abcdwxyz}\color{blue}{\text{product rule of imaginary }e{\text{-powers}}} \\ &=\cos(n\,\varphi)+\ii\,\sin(n\,\varphi)\\ &\phantom{abcdwxyz} \color{blue}{\text{Euler's formula}} \end{aligned}$

$\phantom{x}$

We now give some applications.

Apply de Moivre's formule with $n=2$ : $\begin{aligned}\cos(2\varphi)+\mathrm{i}\,\sin(2\,\varphi) &= (\cos\varphi + \mathrm{i}\sin\varphi)^2 \\ &= \cos^2(\varphi) -\sin^2(\varphi) + \mathrm{i}\cdot 2\cos(\varphi)\sin(\varphi)\end{aligned}$ Thus: $\cos(2\varphi)=\cos^2(\varphi) -\sin^2(\varphi)\qquad\text{and}\qquad \sin(2\varphi)=2\cos(\varphi)\sin(\varphi)$

Apply de Moivre's formule with $n=3$ : $\begin{aligned}\cos(3\varphi)+\mathrm{i}\,\sin(3\,\varphi) &= (\cos\varphi + \mathrm{i}\sin\varphi)^3 \\ &= \bigl(\cos^3(\varphi) -3\cos(\varphi)\sin^2(\varphi)\bigr) \\ &{\phantom=}{} + \mathrm{i}\,\bigl(3\cos^2(\varphi)\sin(\varphi)-\sin^3(\varphi)\bigr)\end{aligned}$ Thus: $\begin{aligned}\cos(3\varphi) &=\cos^3(\varphi) -3\cos(\varphi)\sin^2(\varphi) \\ \sin(3\varphi) &=3\cos^2(\varphi)\sin(\varphi)-\sin^3(\varphi)\end{aligned}$ Because $\cos^2(\varphi)+\sin^2(\varphi)=1$ we can rewrite the last equations as $\begin{aligned}\cos(3\varphi) &=4\cos^3(\varphi) -3\cos(\varphi) \\ \sin(3\varphi) &=-4\sin^3(\varphi)+3\sin(\varphi)\end{aligned}$

The trigonometric sum and difference formulas can also be deduced fairly easyly. An example: $\begin{aligned}\cos(\alpha+\beta)+\mathrm{i}\,\sin(\alpha+\beta) &= e^{\,\mathrm{i}\,(\alpha+\beta)} \\ &= e^{\,\mathrm{i}\,\alpha}\cdot e^{\,\mathrm{i}\,\beta}\\ &= (\cos\alpha + \mathrm{i}\sin\alpha)\cdot (\cos\beta + \mathrm{i}\sin\beta) \\ &= \bigl(\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)\bigr)\\ &{\phantom=}{} + \mathrm{i}\,\bigl(\cos(\alpha)\sin(\beta)+ \sin(\alpha)\cos(\beta)\bigr)\end{aligned}$ Thus: $\begin{aligned} \cos(\alpha+\beta) &=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)\\ \sin(\alpha+\beta) &=\cos(\alpha)\sin(\beta)+ \sin(\alpha)\cos(\beta)\end{aligned}$

For any integer $n$ holds $e^{\,n\,\pi\,\mathrm{i}}=(-1)^n$

de Moivre's formula yields: $\begin{aligned}e^{\,n\,\pi\,\mathrm{i}} &= \cos(n\,\pi)+\mathrm{i}\,\sin(n\,\pi) \\ &= (-1)^n+0\,\mathrm{i} \\ &= (-1)^n\end{aligned}$