Question

Помогите решить комплексные числа
1.(-6-2√3i)^14;

Answer 1

Ответ:

$\frac{48^{7} }{2} +  \sqrt{3} \frac{48^{7} }{2}i$

Пошаговое объяснение:

$z_{1} =(-6-2\sqrt{3} )^{14} \\z_{2} =-6-2\sqrt{3} = x_{2} +iy_{2}  \\r_{2} =\sqrt{x_{2}^{2} +y_{2}^{2}} =\sqrt{36+12} =\sqrt{48} \\x_{2} <0, y_{2} <0 => \alpha_{2} =arctg|\frac{y_{2}}{x_{2}} |-\pi =arctg(\frac{2\sqrt{3}}{6} )-\pi=arctg(\frac{\sqrt{3}}{3} )-\pi=\frac{\pi }{6} -\pi =-\frac{5\pi }{6} \\\\z_{2} = r_{2}e^{ia_{2}} =\sqrt{48}e^{-\frac{5\pi }{6}i}\\z_{1} =z_{2}^{14} =(\sqrt{48})^{14} e^{-\frac{5*14\pi }{6}i}=48^{7} e^{-\frac{35\pi }{3}i}\\\frac{35\pi }{3}i=12\pi -\frac{\pi }{3}i$

$=> z_{1} =48^{7} e^{\frac{\pi }{3}i}\\\alpha_{1}  =\frac{\pi }{3} , r_{1} =48^{7},  \\ 0<\alpha_{1} <\frac{\pi }{2} => x_{1}>0, y_{1}>0 \\=>\frac{y_{1} }{x_{1}}=tg \frac{\pi }{3} =\sqrt{3} => y_{1} =x_{1}\sqrt{3} \\\sqrt{x_{1}^{2}+y_{1}^{2} } =48^{7} =>  x_{1}^{2}+3x_{1}^{2}=48^{14}\\=> x_{1} = \frac{48^{7} }{2} , y_{1} = \sqrt{3} \frac{48^{7} }{2} \\=> z_{1} =  \frac{48^{7} }{2} +  \sqrt{3} \frac{48^{7} }{2}i$