Answer
(a) see proof
(b) The results in (a) do not violate the Conjugate Zeros Theorem because the theorem requires that
all the coefficient of the polynomial be real, a condition not satisfied in this question.
Work Step by Step
(a) $P(2i)=(2i)^2-(1+i)(2i)+2+2i=--4-2i+2+2+2i=0$
$P(1-i)=(1-i)^2-(1+i)(1-i)+2+2i=1-2i-1-(1+1)+2+2i=0$
$P(-2i)=(-2i)^2-(1+i)(-2i)+2+2i=-4+2i-2+2+2i=-4+4i\ne0$
$P(1+i)=(1+i)^2-(1+i)(1+i)+2+2i=2+2i\ne0$
(b) The results in (a) do not violate the Conjugate Zeros Theorem because the theorem requires that
all the coefficient of the polynomial be real, a condition not satisfied in this question.
