Show that 2i and 1-i are both solutions to the equation x^2-(1+i)x+(2+2i)=0 but their conjugates,-2i and 1+i are not. Then explain why this does not violate the Conjugate Zeros Theorem.
The conjugate zeroes theorem says:
"If f(x) is a polynomial having only real coefficients and if a + bi is a zero of f(x), then a – bi is also a zero of f(x)."
Does the given equation satisfy all the conditions required for the theorem to apply?
Looking for the answer myself. I Have not found any help on the internet or YouTube to the solution to this question. Go figure.
So basically the solution is that this property holds "If f(x) is a polynomial having only real coefficients and if a + bi is a zero of f(x), then a – bi is also a zero of f(x)." but only in case, we have all real coefficients. In our case we have (1+i) and (2 + 2i) as imaginary coefficients so the property above does not apply. That's the answer. You can check it by putting roots inside.
To show that 2i and 1-i are solutions to the equation x^2-(1+i)x+(2+2i)=0, we can simply substitute them into the equation and see if it holds true.
For 2i:
Replacing x with 2i in the equation, we get:
(2i)^2 - (1+i)(2i) + (2+2i) = -4 - 2i - 2i + 1 + i + 2 + 2i = -3 + i = 0
Since the result is zero, 2i is indeed a solution.
For 1-i:
Replacing x with 1-i in the equation, we get:
(1-i)^2 - (1+i)(1-i) + (2+2i) = 1 - 2i + i^2 - (1+i) + 1 - i + 2 + 2i = 2 - 2i = 0
Again, the result is zero, so 1-i is also a solution.
Now, let's check the conjugates, -2i and 1+i, to see if they are solutions.
For -2i:
Replacing x with -2i in the equation, we get:
(-2i)^2 - (1+i)(-2i) + (2+2i) = -4i^2 +2i + 2i + 1 - i - 2 + 2i = -4 - 1 + i ≠ 0
The result is not zero, so -2i is not a solution.
For 1+i:
Replacing x with 1+i in the equation, we get:
(1+i)^2 - (1+i)(1+i) + (2+2i) = 1 + 2i + i^2 - 1 - i - i - 1 - 1 + 2i ≠ 0
The result is not zero, so 1+i is also not a solution.
This does not violate the Conjugate Zeros Theorem. The Conjugate Zeros Theorem states that if a polynomial has real coefficients, then any complex solutions must come in conjugate pairs. In this case, the polynomial has complex coefficients, so the Conjugate Zeros Theorem does not apply. Therefore, it is possible for only some of the roots to be conjugates, while others are not. In this case, 2i and 1-i are solutions and their conjugates -2i and 1+i are not solutions.