Soit \(z=a+bi\) .  D'une part, on a

\(\begin{array}{rcl} |z^2|=|(a+bi)^2|&=&|a^2+2abi-b^2|\\ &=&|(a^2-b^2)+2abi|\\ &=&\sqrt{(a^2-b^2)^2+(2ab)^2}\\ &=&\sqrt{a^4-2a^2b^2+b^4+4a^2b^2}\\ &=&\sqrt{a^4+2a^2b^2+b^4}\\ &=&\sqrt{(a^2+b^2)^2}\\ &=&a^2+b^2 \end{array}\)

d'autre part,

\(\begin{array}{rcl} |z|^2&=&|a+bi|^2\\ &=&(\sqrt{a^2+b^2})^2\\ &=&a^2+b^2 \end{array}\)

On a

\(\begin{array}{rcl} z_1\cdot z_2&=&|z_1||z_2|(\cos \phi_1+i\sin \phi_1)(\cos \phi_2+i\sin \phi_2)\\ &=&|z_1||z_2|(\cos \phi_1 \cos \phi_2-\sin \phi_1 \sin \phi_2+i\cos \phi_1\sin \phi_2+i\sin \phi_1\cos \phi_2)\\ &=&|z_1||z_2|(\cos (\phi_1+\phi_2)+i\sin (\phi_1+\phi_2)) \end{array}\)

On a

\(\begin{array}{rcl} \dfrac{z_1}{z_2}&=&\dfrac{|z_1|(\cos \phi_1+i\sin \phi_1)}{|z_2|(\cos \phi_2+i\sin \phi_2)}\\ &=&\dfrac{|z_1|}{|z_2|}\dfrac{(\cos \phi_1+i\sin \phi_1)(\cos \phi_2-i\sin \phi_2)}{(\cos \phi_2+i\sin \phi_2)(\cos \phi_2-i\sin \phi_2)}\\ &=&\dfrac{|z_1|}{|z_2|}\dfrac{\cos \phi_1\cos \phi_2+\sin \phi_1\sin \phi_2+i\sin \phi_1\cos \phi_2-i\cos \phi_1\sin \phi_2}{\cos^2 \phi_2+\sin^2 \phi_2}\\ &=&\dfrac{|z_1|}{|z_2|}(\cos(\phi_1-\phi_2)+i\sin(\phi_1-\phi_2)) \end{array}\)

Démonstration de la formule de Moivre par récurrence :

Si \(n=1\), on a \((\cos \phi+i\sin \phi)^1=\;\cos \phi+i\sin \phi\) .

Supposons que la formule soit vraie pour \(n=p\) et démontrons qu'elle est vraie pour \(n=p+1\).

Hypothèse : \((\cos \phi+i\sin \phi)^p=\;(\cos (p\phi)+i\sin (p\phi))\)

Thèse : \((\cos \phi+i\sin \phi)^{p+1}=\;\cos((p+1) \phi)+i\sin((p+1)\phi)\)

Démonstration :

\( \begin{array}{rcl} (\cos \phi+i\sin \phi)^{p+1}&=&(\cos \phi+i\sin \phi)^{p}.(\cos \phi+i\sin \phi)\\ &=&(\cos (p\phi)+i\sin (p\phi)).(\cos \phi+i\sin \phi)\\ &=&(\cos \phi+i\sin \phi)(\cos (p\phi)+i\sin (p\phi))\\ &=&\cos \phi.\cos (p\phi)+i\cos \phi.\sin (p\phi)+i\sin \phi.\cos (p\phi)-\sin \phi.\sin (p\phi)\\ &=&(\cos \phi.\cos (p\phi)-\sin \phi.\sin (p\phi))+i(\cos \phi.\sin (p\phi)+\sin \phi.\cos (p\phi))\\ &=&\cos (\phi+p\phi)+i\sin (\phi+p\phi)\\ &=&\cos ((p+1)\phi)+i\sin ((p+1)\phi) \end{array}\)