Autour des racines n-ième des nombres complexes - Exercice 5

L'équation à résoudre s'écrit encore :
$$z^5 = \sqrt{3} + i$$
Or :
$$\sqrt{3} + i = \sqrt{\sqrt{3}^2 + 1^2} \times \left( \dfrac{\sqrt{3}}{\sqrt{\sqrt{3}^2 + 1^2}} + i \, \dfrac{1}{\sqrt{\sqrt{3}^2 + 1^2}}\right) = 2 \times \left( \dfrac{\sqrt{3}}{2} + i \, \dfrac{1}{2}\right) = 2 \left( \cos\left( \dfrac{\pi}{6} \right) + i \, \sin\left( \dfrac{\pi}{6} \right) \right) = 2\, e^{i \frac{\pi}{6}}$$
Les cinq solutions complexes $$z_k$$ de l'équation proposée $$z^5 = \sqrt{3} + i$$, avec $$k = \left\{ 0\,;\,1\,;\,2\,;\,3\,;\,4 \right\}$$, sont donc données par :
$$z_k = \sqrt[5]{2} \, e^{i \frac{\frac{\pi}{6}+2k\pi}{5}} = \sqrt[5]{2} \, e^{i \left( \frac{\pi}{30} + \frac{2k\pi}{5} \right)}$$
Nous recherchons donc les racines cinquièmes du nombre complexe $$a=\sqrt{3} + i$$. On a alors les solutions suivantes :
$${\color{blue}{\,\,\, \bullet \,\, \mathrm{si} \,\, k = 0}}$$
Alors :
$$z_0 = \sqrt[5]{2} \, e^{i \frac{\pi}{30}}$$
$${\color{blue}{\,\,\, \bullet \bullet \,\, \mathrm{si} \,\, k = 1}}$$
Alors :
$$z_1 = \sqrt[5]{2} \, e^{i \left( \frac{\pi}{30} + \frac{2\times 1 \times \pi}{5} \right)} = \sqrt[5]{2} \, e^{i \left( \frac{\pi}{30} + \frac{2\pi}{5} \right)} = \sqrt[5]{2} \, e^{i \left( \frac{\pi}{30} + \frac{12\pi}{30} \right)} = \sqrt[5]{2} \, e^{i \left( \frac{\pi + 12 \pi}{30} \right)} = \sqrt[5]{2} \, e^{i \frac{13 \pi}{30}}$$
$${\color{blue}{\,\,\, \bullet \bullet \bullet \,\, \mathrm{si} \,\, k = 2}}$$
Alors :
$$z_2 = \sqrt[5]{2} \, e^{i \left( \frac{\pi}{30} + \frac{2\times 2 \times \pi}{5} \right)} = \sqrt[5]{2} \, e^{i \left( \frac{\pi}{30} + \frac{4\pi}{5} \right)} = \sqrt[5]{2} \, e^{i \left( \frac{\pi}{30} + \frac{24\pi}{30} \right)} = \sqrt[5]{2} \, e^{i \left( \frac{\pi + 24 \pi}{30} \right)} = \sqrt[5]{2} \, e^{i \frac{25 \pi}{30}} = \sqrt[5]{2} \, e^{i \frac{5 \times 5 \times \pi}{5 \times 6}} = \sqrt[5]{2} \, e^{i \frac{5\pi}{6}}$$
$${\color{blue}{\,\,\, \bullet \bullet \bullet \bullet\,\, \mathrm{si} \,\, k = 3}}$$
Alors :
$$z_3 = \sqrt[5]{2} \, e^{i \left( \frac{\pi}{30} + \frac{2\times 3 \times \pi}{5} \right)} = \sqrt[5]{2} \, e^{i \left( \frac{\pi}{30} + \frac{6\pi}{5} \right)} = \sqrt[5]{2} \, e^{i \left( \frac{\pi}{30} + \frac{36\pi}{30} \right)} = \sqrt[5]{2} \, e^{i \left( \frac{\pi + 36 \pi}{30} \right)} = \sqrt[5]{2} \, e^{i \frac{37 \pi}{30}} = \sqrt[5]{2} \, e^{i \frac{(60 - 23) \pi}{30}} = \sqrt[5]{2} \, e^{i \left( 2\pi - \frac{23\pi}{30} \right)}$$
Soit :
$$z_3 = \sqrt[5]{2} \, e^{i \, 2\pi} e^{-i \, \frac{23\pi}{30}} = \sqrt[5]{2} \, e^{i \, 0} e^{-i \, \frac{23\pi}{30}} = \sqrt[5]{2} \, \times 1 \times e^{-i \, \frac{23\pi}{30}} = \sqrt[5]{2} \, e^{-i \, \frac{23\pi}{30}}$$
$${\color{blue}{\,\,\, \bullet \bullet \bullet \bullet \bullet \,\, \mathrm{si} \,\, k = 4}}$$
Alors :
$$z_4 = \sqrt[5]{2} \, e^{i \left( \frac{\pi}{30} + \frac{2\times 4 \times \pi}{5} \right)} = \sqrt[5]{2} \, e^{i \left( \frac{\pi}{30} + \frac{8\pi}{5} \right)} = \sqrt[5]{2} \, e^{i \left( \frac{\pi}{30} + \frac{48\pi}{30} \right)} = \sqrt[5]{2} \, e^{i \left( \frac{\pi + 48 \pi}{30} \right)} = \sqrt[5]{2} \, e^{i \frac{49 \pi}{30}} = \sqrt[5]{2} \, e^{i \frac{(60 - 11) \pi}{30}} = \sqrt[5]{2} \, e^{i \left( 2\pi - \frac{11\pi}{30} \right)}$$
Soit :
$$z_4 = \sqrt[5]{2} \, e^{i \, 2\pi} e^{-i \, \frac{11\pi}{30}} = \sqrt[5]{2} \, e^{i \, 0} e^{-i \, \frac{11\pi}{30}} = \sqrt[5]{2} \, \times 1 \times e^{-i \, \frac{11\pi}{30}} = \sqrt[5]{2} \, e^{-i \, \frac{11\pi}{30}}$$
Finalement, les solutions recherchées sont :
$${\color{red}{\boxed{ z = \sqrt[5]{2} \, e^{i \, \frac{\pi}{30}} \,\,\, \mathrm{ou} \,\,\, z = \sqrt[5]{2} \, e^{i \, \frac{13\pi}{30}} \,\,\, \mathrm{ou} \,\,\, z = \sqrt[5]{2} \, e^{i \, \frac{5\pi}{6}} \,\,\, \mathrm{ou} \,\,\, z = \sqrt[5]{2} \, e^{-i \, \frac{23\pi}{30}} \,\,\, \mathrm{ou} \,\,\, z = \sqrt[5]{2} \, e^{-i \, \frac{11\pi}{30}}}}}$$