Fórmula binómica y trigonométrica de números complejos

9000035805

Parte: 
B
Dados los números complejos \[ \text{$a = 2\left (\cos \frac{2\pi } {3} + \mathrm{i}\sin \frac{2\pi } {3}\right )$, $b = \sqrt{2}\left (\cos \frac{3\pi } {4} + \mathrm{i}\sin \frac{3\pi } {4}\right )$,} \] determina el producto \(ab\).
\(2\sqrt{2}\left (\cos \frac{17\pi } {12} + \mathrm{i}\sin \frac{17\pi } {12}\right )\)
\(2\sqrt{2}\left (\cos \frac{\pi }{2} + \mathrm{i}\sin \frac{\pi }{2}\right )\)
\(2\sqrt{2}\left (\cos \frac{5\pi } {7} + \mathrm{i}\sin \frac{5\pi } {7}\right )\)
\(2\sqrt{2}\left (\cos \frac{5\pi } {12} + \mathrm{i}\sin \frac{5\pi } {12}\right )\)

9000035806

Parte: 
B
Dados los números complejos \[ \text{ $a = 2\left (\cos \frac{5\pi } {3} + \mathrm{i}\sin \frac{5\pi } {3}\right )$, $b = 3\left (\cos \frac{11\pi } {6} + \mathrm{i}\sin \frac{11\pi } {6} \right )$,} \] determina el cociente \(\frac{a} {b}\).
\(\frac{2} {3}\left (\cos \frac{11\pi } {6} + \mathrm{i}\sin \frac{11\pi } {6} \right )\)
\(\frac{2} {3}\left (\cos \frac{\pi } {6} + \mathrm{i}\sin \frac{\pi } {6}\right )\)
\(\frac{2} {3}\left (\cos \frac{5\pi } {6} + \mathrm{i}\sin \frac{5\pi } {6}\right )\)
\(\frac{2} {3}\left (\cos \frac{7\pi } {6} + \mathrm{i}\sin \frac{7\pi } {6}\right )\)

9000034807

Parte: 
B
Determina la forma polar del número complejo \(z = 2\mathrm{i}\).
\(2\left (\cos \frac{\pi }{2} + \mathrm{i}\sin \frac{\pi }{2}\right )\)
\(\sqrt{2}\left (\cos \frac{\pi }{2} + \mathrm{i}\sin \frac{\pi }{2}\right )\)
\(\cos \frac{\pi }{2} + \mathrm{i}\sin \frac{\pi }{2}\)
\(2\left (\cos 0 + \mathrm{i}\sin 0\right )\)

9000034809

Parte: 
B
Dados los números complejos \(z_{1} = 2\left (\cos \frac{\pi }{6} + \mathrm{i}\sin \frac{\pi }{6}\right )\) y \(z_{2} = \sqrt{3}\left (\cos \frac{4\pi } {3} + \mathrm{i}\sin \frac{4\pi } {3}\right )\), determina el ángulo del producto \(z_{1}z_{2}\) en forma polar.
\(\frac{3\pi } {2}\)
\(\frac{2} {9}\pi \)
\(\frac{5} {9}\pi \)
\(3\pi \)

9000034810

Parte: 
B
Dados los números complejos \(z_{1} = 2\left (\cos \frac{\pi }{4} + \mathrm{i}\sin \frac{\pi }{4}\right )\) y \(z_{2} = \sqrt{2}\left (\cos \frac{7\pi } {4} + \mathrm{i}\sin \frac{7\pi } {4}\right )\), determina el ángulo en la forma polar del cociente \(\frac{z_{1}} {z_{2}} \).
\(\frac{\pi } {2}\)
\(- \frac{\pi } {2}\)
\(-\frac{3} {2}\pi \)
\(\frac{3} {2}\pi \)