How do i find P(2)?
Let $P(x)$ be the polynomial of least degree with real coefficients such that $P(i)=P(1+i)=0$ and $P(1)=4$. Find $P(2)$. Here i is the complex number. I have done such questions with real numbers but I don't know why I am not getting it.
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$\begingroup$The condition $P(i)=0$ implies $x^2+1\mid P(x)$, while $P(1+i)=0$ implies $(x-1)^2+1\mid P(x)$, thus $P(x)=a(x^2+1)(x^2-2x+2)$ for some $a\in\mathbb R$, because $x^2+1$ and $(x-1)^2+1=x^2-2x+2$ are relatively prime. Since $P(1)=4$ we get $4=P(1)=4a$, hence $a=1$, thus $P(x)=(x^2+1)(x^2-2x+2)$ and $P(2)=10$.
$\endgroup$ $\begingroup$Hint: $P$ has real coefficients. So, if $P(i) = 0$, then $P(-i) = 0$. If $P(1+i) = 0$, then $P(1 - i) = 0$.
Can you construct the smallest complex polynomial that has these zeros?
$\endgroup$ $\begingroup$Some hints:
1) When a polynomial has real coefficients the complex roots always come in conjugate pairs, i.e. if $z = a+ib$ is a root, then $\overline{z} = a-ib$ is root as well.
2) A polynomial of degree $n$ can be written on the form $$p(x)=\alpha\prod\limits_{k=1}^n(x-r_k) $$ where $r_k$ are the roots and $\alpha$ is the coefficient of the term with highest degree.
Example: A polynomial of degree 4 can be written as
$$p(x) = \alpha(x-r_1)(x-r_2)(x-r_3)(x-r_4) $$
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