Given the equation x8−1=0 Because equation degree is equal to = 8 - contains the even number 8 in the numerator, then the equation has two real roots. Get the root 8-th degree of the equation sides: We get: 8x8=81 8x8=(−1)81 or x=1 x=−1 We get the answer: x = 1 We get the answer: x = -1 or x1=−1 x2=1
All other 6 root(s) is the complex numbers. do replacement: z=x then the equation will be the: z8=1 Any complex number can presented so: z=reip substitute to the equation r8e8ip=1 where r=1 - the magnitude of the complex number Substitute r: e8ip=1 Using Euler’s formula, we find roots for p isin(8p)+cos(8p)=1 so cos(8p)=1 and sin(8p)=0 then p=4πN where N=0,1,2,3,... Looping through the values of N and substituting p into the formula for z Consequently, the solution will be for z: z1=−1 z2=1 z3=−i z4=i z5=−22−22i z6=−22+22i z7=22−22i z8=22+22i do backward replacement z=x x=z
The final answer: x1=−1 x2=1 x3=−i x4=i x5=−22−22i x6=−22+22i x7=22−22i x8=22+22i