How do you find all the real and complex roots of z^5 + 1 = 0z5+1=0?
1 Answer
Explanation:
0 = z^5+10=z5+1
color(white)(0) = (z+1)(z^4-z^3+z^2-z+1)0=(z+1)(z4−z3+z2−z+1)
color(white)(0) = (z+1)z^2(z^2-z+1-1/z+1/z^2)0=(z+1)z2(z2−z+1−1z+1z2)
color(white)(0) = (z+1)z^2((z+1/z)^2+(z+1/z)-1)0=(z+1)z2((z+1z)2+(z+1z)−1)
color(white)(0) = (z+1)z^2(((z+1/z)+1/2)^2-(sqrt(5)/2)^2)0=(z+1)z2⎛⎝((z+1z)+12)2−(√52)2⎞⎠
color(white)(0) = (z+1)z^2(z+1/z+1/2-sqrt(5)/2)(z+1/z+1/2+sqrt(5)/2)0=(z+1)z2(z+1z+12−√52)(z+1z+12+√52)
color(white)(0) = (z+1)(z^2+(1/2-sqrt(5)/2)z+1)(z^2+(1/2+sqrt(5)/2)z+1)0=(z+1)(z2+(12−√52)z+1)(z2+(12+√52)z+1)
So:
0 = z+1 rarr z = -10=z+1→z=−1
or:
0 = z^2+(1/2-sqrt(5)/2)z+10=z2+(12−√52)z+1
and hence:
z = 1/2(-1/2+sqrt(5)/2+-sqrt((1/2-sqrt(5)/2)^2-4))z=12⎛⎜⎝−12+√52± ⎷(12−√52)2−4⎞⎟⎠
color(white)(z) = 1/4(sqrt(5)-1)+-1/4sqrt(10+2sqrt(5))iz=14(√5−1)±14√10+2√5i
or:
0 = z^2+(1/2+sqrt(5)/2)z+10=z2+(12+√52)z+1
and hence:
z = 1/2(-1/2-sqrt(5)/2+-sqrt((1/2+sqrt(5)/2)^2-4))z=12⎛⎜⎝−12−√52± ⎷(12+√52)2−4⎞⎟⎠
color(white)(z) = 1/4(-sqrt(5)-1)+-1/4sqrt(10-2sqrt(5))iz=14(−√5−1)±14√10−2√5i
