x $ ^6-i=0 \\ \; \\ \; \\ \therefore x^6 \;=\; i \; = \; cos\dfrac{\pi}{2}+i sin\dfrac{\pi}{2} \; = \; cos\bigg(2k\pi +\dfrac{\pi}{2}) +i \;sin\bigg(2k\pi +\dfrac{\pi}{2}) \\ \; \\ \; \\ \therefore x^6 \;=\; cos(4k+1)\dfrac{\pi}{2} \;+\; i sin(4k+1)\dfrac{\pi}{2} \; \; \; \; or \\ \; \\ \; \\ \therefore x \;=\; \Bigg[ cos(4k+1)\dfrac{\pi}{2} \;+\; i sin(4k+1)\dfrac{\pi}{2} \Bigg]^{\frac{1}{6}} \; \; \; \; k=0,1,2,3,4,5 \\ \; \\ \; \\ \therefore x \;=\; \Bigg[ cos(4k+1)\dfrac{\pi}{12} \;+\; i sin(4k+1)\dfrac{\pi}{12} \Bigg] \; \; \; \; \; \; \\ \{ By \; De-Moivre’s \; theorem, (cos\theta+isin\theta)^n \;=\; cosn\theta \;+\; isin\;n\theta \} \\ \; \\ \; \\ \; \\ \; \\ x_0 \; = \; cos \dfrac{\pi}{12} \; + \; isin \dfrac{\pi}{12} \\ \; \\ \; \\ \; \\ x_1 \; = \; cos \dfrac{5\pi}{12} \; + \; isin \dfrac{5\pi}{12} \\ \; \\ \; \\ \; \\ x_2 \; = \; cos \dfrac{9\pi}{12} \; + \; isin \dfrac{9\pi}{12} \; = \; cos \dfrac{3\pi}{4} \; + \; isin \dfrac{3\pi}{4} \\ \; \\ \; \\ \; \\ x_3 \; = \; cos \dfrac{13\pi}{12} \; + \; isin \dfrac{13\pi}{12} \\ \; \\ \; \\ \; \\ x_4 \; = \; cos \dfrac{17\pi}{12} \; + \; isin \dfrac{17\pi}{12} \\ \; \\ \; \\ \; \\ x_5 \; = \; cos \dfrac{21\pi}{12} \; + \; isin \dfrac{21\pi}{12} \;=\; cos \dfrac{7\pi}{4} \; + \; isin \dfrac{7\pi}{4} $

Thus, we get the no. of roots that is equal to power of the equation, i.e. 6