Fórmula binómica y trigonométrica de números complejos

2010013102

Parte: 
B
Dados los números complejos \( a=2\left(\cos⁡ \frac{\pi}{3}+\mathrm{i}\sin⁡ \frac{\pi}{3}\right) \), \( b=\sqrt{2}\left(\cos⁡ \frac{5\pi}{4}+\mathrm{i}\sin \frac{5\pi}{4}\right) \) y \( c=2\sqrt{2}\left(\cos \left(-\frac{\pi}{6}\right)+\mathrm{i}\sin \left(-\frac{\pi}{6}\right)\right) \), calcula \( a\cdot b\cdot c \).
\(8\left(\cos \frac{17\pi}{12}+\mathrm{i}\sin \frac{17\pi}{12} \right) \)
\(8\left(\cos \frac{17\pi}{12}-\mathrm{i}\sin \frac{17\pi}{12} \right) \)
\(8\left(\cos \frac{7\pi}{4}+\mathrm{i}\sin \frac{7\pi}{4} \right) \)
\(4\sqrt{2}\left(\cos \frac{17\pi}{12}+\mathrm{i}\sin \frac{17\pi}{12} \right) \)

2010010403

Parte: 
A
Dados los números complejos \[ a = \sqrt{2} + \mathrm{i},\ \quad b = {2} -\sqrt{3}\mathrm{i},\ \] determina \(\frac{a} {b}\).
\(\frac{2\sqrt{2}-\sqrt{3}} {7} + \mathrm{i}\frac{\sqrt{6}+2} {7} \)
\(\frac{2\sqrt{2}-\sqrt{3}} {7} - \mathrm{i}\frac{\sqrt{6}+2} {7} \)
\(\frac{2\sqrt{2}+\sqrt{3}} {4} + \mathrm{i}\frac{\sqrt{6}-2} {4} \)
\(\frac{\sqrt{2}-\sqrt{6}} {3} + \mathrm{i}\frac{\sqrt{3}+2} {3} \)