Nombres complexes

Pour \(\underline{z} = \cos x + j \sin x\) on a : \(\underline{z}^{\ast} = \cos x - j \sin x\) ; \(\cos x = \frac{\underline{z} + \underline{z}^{\ast}}{2}\); \(\sin x = \frac{\underline{z}-\underline{z}^{\ast}}{2j}\) et \(\underline{z}~\underline{z}^{\ast} = 1\)

donc :

\(E(x) = \sin^{3} x ~\cos^{2} x = - \frac{\left(\underline{z} - \underline{z}^{\ast}\right)^{3}}{8j} \frac{\left(\underline{z} + \underline{z}^{\ast}\right)^{2}}{4} = -\frac{1}{32 j} \left(\underline{z}^{2} - {\underline{z}^{\ast}}^{2}\right)^{2} \left(\underline{z} - \underline{z}^{\ast}\right) = -\frac{1}{32j}\left(\underline{z}^{4} - 2 + {\underline{z}^{\ast}}^{4}\right) \left(\underline{z} - \underline{z}^{\ast}\right)\)

\(\Leftrightarrow -32 j \sin^{3} x \cos^{2} x = \left(\underline{z}^{5}- {\underline{z}^{\ast}}^{5} - 2\left(\underline{z} - \underline{z}^{\ast}\right) - \left(\underline{z}^{3} -{\underline{z}^{\ast}}^{3}\right)\right)\)

\(\Rightarrow \sin^{3}x \cos^{2} x = \frac{1}{16} \left(- \sin 5x + \sin 3x + 2 \sin x\right)\) ( 6 points )