Seja o número complexo:
[tex]\boxed{z=a+bi}\ \therefore\ \boxed{z=-1}\ \therefore\ (a,b)=(-1,0)[/tex]
Podemos escrevê-lo em sua forma polar:
[tex]\boxed{z=\rho(\cos{\theta}+i\sin{\theta})}[/tex]
[tex]\rho=\sqrt{a^2+b^2}=\sqrt{(-1)^2+(0)^2}=\sqrt{1}\ \therefore\ \rho=1[/tex]
[tex]\cos{\theta}=\dfrac{a}{\rho}=\dfrac{-1}{1}\ \therefore\ \cos{\theta}=-1[/tex]
[tex]\sin{\theta}=\dfrac{b}{\rho}=\dfrac{0}{1}\ \therefore\ \sin{\theta}=0\ \therefore\ \boxed{\theta=\pi}[/tex]
[tex]z=1(\cos{\pi}+i\sin{\pi})\ \therefore\ \boxed{z=\cos{\pi}+i\sin{\pi}}[/tex]
Dessa forma, podemos utilizar a Fórmula de De Moivre para calcular [tex]z^{\tfrac{7}{8}}[/tex]:
[tex]\boxed{z^n=\rho^n(\cos{n\theta}+i\sin{n\theta})}[/tex]
[tex](-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}}}[/tex]
O [tex]\sin{\frac{7\pi}{8}}[/tex] e o [tex]\cos{\frac{7\pi}{8}}[/tex] podem ser calculados através de manipulações trigonométricas com as fórmulas de soma de arcos e arco metade:
[tex]\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)}=[/tex]
[tex]\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}}=[/tex]
[tex]\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}}[/tex]
[tex]\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[/tex]
[tex]\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}}[/tex]
Dessa forma, a expressão será:
[tex]\boxed{z^{\tfrac{7}{8}}=\dfrac{\sqrt{2+\sqrt{2}}}{2}-\dfrac{\sqrt{2-\sqrt{2}}}{2}i}[/tex]
[tex]\boxed{(-1)^{\tfrac{7}{8}}\approx0.92388-0.3827i}[/tex]
