Examples
The following numbers are algebraic:
3,
e2
i/5,
2cos(2/5) = e2i/5
+ e - 2i/5 =
(-1 + 5)/2.
Clearly, it is a zero of
X2 - 3 = 0.
Clearly, it is a zero of
X5 - 1 = 0.
X4 + X3 + X2 + X + 1.
But it is not a zero of the linear factor X - 1, so it is a zero of the quotient:
Put a =
e2i/5.
Then, as we have seen in the previous example,
a4 + a3 + a2 + a + 1 = 0.
(a + a-1)2 - 2 +
(a + a-1) + 1 = 0,
i/5.
Then, as we have seen in the previous example,
Multiply by a-2 and replace a2 + a-2 by (a + a-1)2 - 2. Then we have
from which we conclude that
2cos(2/5) = a + a-1 is a zero of
X2 + X - 1.