Question

Raiz cubica de -64i aplicando moivre

Answer 1

tenemos a z=-64i en la forma binómica, primero hay que llevarlo a su forma polar para poder aplicar la fórmula de moivre para raíces

fórmula de moivre

$z^{\frac{1}{n}}=[r(cos(\alpha)+isen(\alpha)]^{\frac{1}{n} = r^{\frac{1}{n}}[cos(\frac{\alpha+2k\pi}{n})+isen(\frac{\alpha+2k\pi}{n})]$

donde n es la ráiz que neceistamos (3) y 0≤k∈Z≤n-1

en z=-64i, x = 0 e y=-64

$r=\sqrt{x^{2}+y^{2}}=\sqrt{0^{2}+(-64)^{2}}= \sqrt{(-64)^{2}}=64$

$\alpha=\frac{3\pi}{2}$

sustityendo

$z=-64i=64[cos(\frac{3\pi}{2})+isen(\frac{3\pi}{2})]$

como queremos hallar la raíz cúbica n=3, entonces

$z^{\frac{1}{3}}=64^{\frac{1}{3}}[cos(\frac{\frac{3\pi}{2}+2k\pi}{3})+isen(\frac{\frac{3\pi}{2}+2k\pi}{3})]$

entonces las 3 raíces serían

$(64)^{\frac{1}{3}}=4\\z_{0}=4[cos(\frac{\frac{3\pi}{2}+2*0*\pi}{3})+isen(\frac{\frac{3\pi}{2}+2*0*\pi}{3})]=4[cos(\frac{\pi}{2})+isen(\frac{\pi}{2})]=4[0+i*1]=4i\\z_{1}=4[cos(\frac{\frac{3\pi}{2}+2*1*\pi}{3})+isen(\frac{\frac{3\pi}{2}+2*1*\pi}{3})]=4[cos(\frac{\frac{3\pi}{2}+2\pi}{3})+isen(\frac{\frac{3\pi}{2}+2\pi}{3})]=4[cos(\frac{7\pi}{6})+isen(\frac{7\pi}{6})]=4[-\frac{\sqrt(3)}{2}-i\frac{1}{2}]=-2\sqrt(3)-2i$$z_{2}=4[cos(\frac{\frac{3\pi}{2}+2*2*\pi}{3})+isen(\frac{\frac{3\pi}{2}+2*2*\pi}{3})]=4[cos(\frac{\frac{3\pi}{2}+2*2\pi}{3})+isen(\frac{\frac{3\pi}{2}+2*2\pi}{3})]=4[cos(\frac{11\pi}{6})+isen(\frac{11\pi}{6})]=4[\frac{\sqrt(3)}{2}-i\frac{1}{2}]=2\sqrt(3)-2i$

saludos..