A complex number a≠-1 is the root of an equation x
3+1=0.
Find 1+2a+3a2+4a3+5a4.
Use substitution a^3 = -1
1 + 2a + 3a^2 + 4a^3 + 5a^4
1 + 2a + 3a^2 - 4 - 5a
-3 - 3a + 3a^2
= 3(a^2 - a - 1)
The 3 cube roots of -1 are at 60, 180, and 270 degrees on the unit circle in the complex plane.
r1 = .5 + .5*√3i
r2 = .5 - .5*√3i
r3 = -1 but this is excluded
(r1)^2 = .25 - .75 + .5*√3i = -.5 + .5*√3i
(r1)^2 - (r1) - 1 = (-.5 + .5*√3i) - (.5 + .5*√3i) - 1
= -2
3(a^2 - a - 1) = -6
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Posted by Larry
on 2024-04-23 10:32:00 |
Another way
