Write the equation of lowest degree with real coefficients if two of its roots are -1 and 1+i.
The answer is x^3-x^2+2=0. I don't understand why.
I did this on paper first, but then messed up when I typed it.
the first few lines in the equations should have been:
(x+1)(x - (1+i))(x - (1-i)) = 0
(x+1)(x-1-i)(x-1+i) = 0
(x+1)(x^2 - x + xi -x +1 -i -xi + i - i^2) = 0
the rest is fine
complex roots always come in conjugate pairs, so if one is 1+i, you must have another as 1-i
so we know the equation is
(x+1)(x - (1+i))(x + 1+i) = 0
(x+1)(x-1-i)(x+1+i) = 0
(x+1)(x^2 + x + xi -x -x -i -xi - i - x^2) = 0
(x+1)(x^2 - 2x + 2) = 0
x^3 - x^2 + 2x + x^2 - 2x + 2 = 0
x^3 - x^2 + 2=0
To find the equation of lowest degree with real coefficients when given two roots, we need to determine the complex conjugate of the complex root. In this case, the complex conjugate of 1+i is 1-i.
Let's call the third root of the equation, which is the complex conjugate of 1+i, as a.
Therefore, the equation can be written in factored form as follows:
(x - (-1))(x - (1 + i))(x - (1 - i)) = 0
Simplifying this equation, we get:
(x + 1)(x - 1 - i)(x - 1 + i) = 0
Expanding this equation further, we get:
(x + 1)(x^2 - 1 + xi - 1 - xi + i^2) = 0
Since i^2 = -1, we can simplify this equation as:
(x + 1)(x^2 - 2 + i^2) = 0
(x + 1)(x^2 - 2 - 1) = 0
(x + 1)(x^2 - 3) = 0
Expanding this equation:
x^3 - 3x + x^2 - 3 = 0
Rearranging the terms:
x^3 + x^2 - 3x - 3 = 0
Therefore, the equation of the lowest degree with real coefficients, given that the roots are -1 and 1+i, is:
x^3 + x^2 - 3x - 3 = 0.
Note: There could be multiple correct answers, but this is one possible equation.
To determine the equation of lowest degree with real coefficients, we need to consider the complex conjugate of the given complex root. The complex conjugate of 1+i is 1-i. Therefore, the three roots of the equation are -1, 1+i, and 1-i.
To find the equation, we can start by using the fact that the complex roots of a polynomial equation with real coefficients always occur in conjugate pairs.
The equation, in factored form, can be written as:
(x - (-1))(x - (1+i))(x - (1-i)) = 0
Simplifying this expression, we have:
(x + 1)(x - 1 - i)(x - 1 + i) = 0
Expanding the product of the binomials, we get:
(x + 1)(x^2 - (1 - i)x - (1 + i)x + (1 - i)(1 + i)) = 0
(x + 1)(x^2 - (x - ix) - (x + ix) + (1 - i^2)) = 0
(x + 1)(x^2 - 2x + (1 + 1)) = 0
(x + 1)(x^2 - 2x + 2) = 0
Now, if we multiply out the expression, we get:
x^3 - 2x^2 + 2x + x^2 - 2x + 2 = 0
x^3 - x^2 + 2x + 2 = 0
Therefore, the equation of lowest degree with real coefficients and with roots -1 and 1+i is x^3 - x^2 + 2x + 2 = 0.