Ejercicios resueltos de conversión rectangular-polar

$r_1 = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$

$\theta_1 = \arctan\left(\frac{3}{3}\right) = \arctan(1) = \frac{\pi}{4}$ (primer cuadrante)

Forma polar: $z_1 = 3\sqrt{2}e^{i\pi/4}$

$r_2 = \sqrt{(-4)^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8$

$\theta_2 = \arctan\left(\frac{4\sqrt{3}}{-4}\right) + \pi = \arctan(-\sqrt{3}) + \pi = -\frac{\pi}{3} + \pi = \frac{2\pi}{3}$ (segundo cuadrante)

Forma polar: $z_2 = 8e^{i2\pi/3}$

$r_3 = \sqrt{0^2 + (-5)^2} = 5$

$\theta_3 = \frac{3\pi}{2}$ (eje imaginario negativo)

Forma polar: $z_3 = 5e^{i3\pi/2}$

$r_4 = 6$

$\theta_4 = 0$ (eje real positivo)

Forma polar: $z_4 = 6e^{i0}$

$a_1 = 2\cos\left(\frac{\pi}{3}\right) = 2 \cdot \frac{1}{2} = 1$

$b_1 = 2\sin\left(\frac{\pi}{3}\right) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$

Forma rectangular: $z_1 = 1 + \sqrt{3}i$

$a_2 = 5\cos\left(\frac{5\pi}{4}\right) = 5 \cdot \left(-\frac{\sqrt{2}}{2}\right) = -\frac{5\sqrt{2}}{2}$

$b_2 = 5\sin\left(\frac{5\pi}{4}\right) = 5 \cdot \left(-\frac{\sqrt{2}}{2}\right) = -\frac{5\sqrt{2}}{2}$

Forma rectangular: $z_2 = -\frac{5\sqrt{2}}{2} - \frac{5\sqrt{2}}{2}i$

$a_3 = 3\cos\left(\frac{2\pi}{3}\right) = 3 \cdot \left(-\frac{1}{2}\right) = -\frac{3}{2}$

$b_3 = 3\sin\left(\frac{2\pi}{3}\right) = 3 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}$

Forma rectangular: $z_3 = -\frac{3}{2} + \frac{3\sqrt{3}}{2}i$

$a_4 = 4\cos\left(-\frac{\pi}{6}\right) = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}$

$b_4 = 4\sin\left(-\frac{\pi}{6}\right) = 4 \cdot \left(-\frac{1}{2}\right) = -2$

Forma rectangular: $z_4 = 2\sqrt{3} - 2i$

Para $z_1 = 1 - \sqrt{3}i$:

$r_1 = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2$

$\theta_1 = \arctan\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3}$ (cuarto cuadrante)

$z_1 = 2e^{-i\pi/3}$

Para $z_2 = \sqrt{2} + \sqrt{2}i$:

$r_2 = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = 2$

$\theta_2 = \arctan\left(\frac{\sqrt{2}}{\sqrt{2}}\right) = \arctan(1) = \frac{\pi}{4}$ (primer cuadrante)

$z_2 = 2e^{i\pi/4}$

$z_1z_2 = 2e^{-i\pi/3} \cdot 2e^{i\pi/4} = 4e^{i(-\pi/3 + \pi/4)} = 4e^{i(-\pi/12)} = 4e^{-i\pi/12}$

$\frac{z_1}{z_2} = \frac{2e^{-i\pi/3}}{2e^{i\pi/4}} = e^{i(-\pi/3 - \pi/4)} = e^{i(-7\pi/12)}$

$z_1z_2 = (1 - \sqrt{3}i)(\sqrt{2} + \sqrt{2}i) = \sqrt{2} + \sqrt{2}i - \sqrt{6}i - \sqrt{6}i^2 = \sqrt{2} + \sqrt{6} + i(\sqrt{2} - \sqrt{6})$

Conversión del resultado polar: $4e^{-i\pi/12} = 4[\cos(-\pi/12) + i\sin(-\pi/12)]$

$= 4\left[\frac{\sqrt{6} + \sqrt{2}}{4} - i\frac{\sqrt{6} - \sqrt{2}}{4}\right] = \sqrt{6} + \sqrt{2} - i(\sqrt{6} - \sqrt{2})$ ✓

$\frac{z_1}{z_2} = \frac{1 - \sqrt{3}i}{\sqrt{2} + \sqrt{2}i} \cdot \frac{\sqrt{2} - \sqrt{2}i}{\sqrt{2} - \sqrt{2}i} = \frac{(\sqrt{2} - \sqrt{6}) - i(\sqrt{6} + \sqrt{2})}{4}$

$= \frac{\sqrt{2} - \sqrt{6}}{4} - i\frac{\sqrt{6} + \sqrt{2}}{4}$

Conversión del resultado polar: $e^{-i7\pi/12} = \cos(-7\pi/12) - i\sin(7\pi/12)$ ✓

$w_1w_2 = re^{i\alpha} \cdot se^{i\beta} = rse^{i(\alpha + \beta)}$

$|w_1w_2| = |rse^{i(\alpha + \beta)}| = rs = |re^{i\alpha}||se^{i\beta}| = |w_1||w_2|$ ✓

$\arg(w_1w_2) = \alpha + \beta = \arg(w_1) + \arg(w_2)$ ✓

$w_1 = 3e^{i\pi/6}$, $w_2 = 2e^{i\pi/4}$

$w_1w_2 = 3 \cdot 2 \cdot e^{i(\pi/6 + \pi/4)} = 6e^{i5\pi/12}$

Verificación: $|w_1w_2| = 6$, $|w_1||w_2| = 3 \cdot 2 = 6$ ✓

$\arg(w_1w_2) = \frac{5\pi}{12}$, $\arg(w_1) + \arg(w_2) = \frac{\pi}{6} + \frac{\pi}{4} = \frac{5\pi}{12}$ ✓

$(w_1w_2)^3 = (6e^{i5\pi/12})^3 = 216e^{i15\pi/12} = 216e^{i5\pi/4}$

$w_1^3w_2^3 = (3e^{i\pi/6})^3(2e^{i\pi/4})^3 = 27e^{i\pi/2} \cdot 8e^{i3\pi/4} = 216e^{i5\pi/4}$ ✓

$w_1w_2 = 6e^{i5\pi/12}$

Las raíces cuadradas son: $\sqrt{6}e^{i5\pi/24}$ y $\sqrt{6}e^{i(5\pi/24 + \pi)} = \sqrt{6}e^{i29\pi/24}$

Verificación: $(\sqrt{6}e^{i5\pi/24})^2 = 6e^{i5\pi/12}$ ✓

$(\sqrt{6}e^{i29\pi/24})^2 = 6e^{i29\pi/12} = 6e^{i(24\pi + 5\pi)/12} = 6e^{i5\pi/12}$ ✓