Question

Calcule o valor de : e) (-1) ⅞

Answer 1

Seja o número complexo:

$\boxed{z=a+bi}\ \therefore\ \boxed{z=-1}\ \therefore\ (a,b)=(-1,0)$

Podemos escrevê-lo em sua forma polar:

$\boxed{z=\rho(\cos{\theta}+i\sin{\theta})}$

$\rho=\sqrt{a^2+b^2}=\sqrt{(-1)^2+(0)^2}=\sqrt{1}\ \therefore\ \rho=1$

$\cos{\theta}=\dfrac{a}{\rho}=\dfrac{-1}{1}\ \therefore\ \cos{\theta}=-1$

$\sin{\theta}=\dfrac{b}{\rho}=\dfrac{0}{1}\ \therefore\ \sin{\theta}=0\ \therefore\ \boxed{\theta=\pi}$

$z=1(\cos{\pi}+i\sin{\pi})\ \therefore\ \boxed{z=\cos{\pi}+i\sin{\pi}}$

Dessa forma, podemos utilizar a Fórmula de De Moivre  para calcular $z^{\tfrac{7}{8}}$:

$\boxed{z^n=\rho^n(\cos{n\theta}+i\sin{n\theta})}$

$(-1)^{\tfrac{7}{8}}=z^{\tfrac{7}{8}}\ \therefore\ \boxed{z^{\tfrac{7}{8}}=\cos{\dfrac{7\pi}{8}}+i\sin{\dfrac{7\pi}{8}}}$

O $\sin{\frac{7\pi}{8}}$ e o $\cos{\frac{7\pi}{8}}$ podem ser calculados através de manipulações trigonométricas com as fórmulas de soma de arcos e arco metade:

$\cos{\dfrac{7\pi}{8}}=\cos{\bigg(\dfrac{4(2\pi-\frac{\pi}{4})}{8}\bigg)}=\cos{\bigg(\dfrac{2\pi-\frac{\pi}{4}}{2}\bigg)}=$

$\sqrt{\dfrac{1+\cos{(2\pi-\frac{\pi}{4})}}{2}}=\sqrt{\dfrac{1+\cos{2\pi}\cos{\frac{\pi}{4}}+\sin{2\pi}\sin{\frac{\pi}{4}}}{2}}=$

$\sqrt{\dfrac{1+\cos{\frac{\pi}{4}}}{2}}=\sqrt{\dfrac{1+\frac{\sqrt{2}}{2}}{2}}\ \therefore\ \boxed{\cos{\dfrac{7\pi}{8}}=\dfrac{\sqrt{2+\sqrt{2}}}{2}}$

$\sin^2{\dfrac{7\pi}{8}}+\cos^2{\dfrac{7\pi}{8}}=1\ \therefore\ \sin^2{\dfrac{7\pi}{8}}=1-\bigg(\dfrac{\sqrt{2+\sqrt{2}}}{2}\bigg)^2\ \therefore$

$\sin^2{\dfrac{7\pi}{8}}=\dfrac{4}{4}-\dfrac{2+\sqrt{2}}{4}=\dfrac{2-\sqrt{2}}{4}\ \therefore\ \boxed{\sin{\dfrac{7\pi}{8}}=-\dfrac{\sqrt{2-\sqrt{2}}}{2}}$

Dessa forma, a expressão será:

$\boxed{z^{\tfrac{7}{8}}=\dfrac{\sqrt{2+\sqrt{2}}}{2}-\dfrac{\sqrt{2-\sqrt{2}}}{2}i}$

$\boxed{(-1)^{\tfrac{7}{8}}\approx0.92388-0.3827i}$