Question: (a) Show that $2i$ and $1 – i$ are both solutions of the equation
$x^2 – (1 + i)x + (2 + 2i) = 0$ but that their complex conjugates $-2i$ and $1 + i$ are not.
(b) Explain why the result of part (a) does not violate the Conjugate Zeros Theorem.
I'm done with both of them and it turns out the reason for the 2nd one not being true is because the theorem is applicable only to the polynomials with real coefficient, but in this case, we've got $2i$ & $1 -i$ which are complex.
Next question: (a) Find the polynomial with real coefficients of the smallest
possible degree for which $i$ and $1 + i$ are zeros and in which the coefficient of the highest power is 1.
(b) Find the polynomial with complex coefficients of the smallest possible degree for which $i$ and $1 + i$ are zeros and in which the coefficient of the highest power is 1.
Why is it that the answers for both the parts of the 2nd question (the 'Next Question') are different? In both the cases, the coefficient (see this), as far as I can tell, are complex. Is it not so?
Best Answer
In the first part of your question you have noted that the complex conjugate roots theorem is applicable only to polynomial with real coefficients. This means that, if the coefficients of the polynomial are real numbers, than the complex roots are couples of conjugate numbers.
So, a polynomial with real coefficients, that has roots $x_1=i$ and $x_2=1+i$, must have olso the roots $x_3=-i$ and $x_4=1-i$ and it is the degree $4$ polynomial: $$ (x-i)(x+i)(x-1-i)(x-1+i)=(x^2+1)(x^2-2x+2)=$$ $$=x^4-2x^3+3x^2-2x+2 \qquad (1) $$
The theorem is not true for a polynomial with complex coefficients, in the sense that a polynomial with complex coefficients can have complex roots that are not conjugate, and this is the case for the polynomial
$$ (x-i)(x-1-i)=x^2-(1+2i)x-1+i\qquad (2) $$
Obviously also $(1)$ can be considered as a polynomial with complex coefficients, thinking to the real numbers as a subset of complex numbers, but note that: