Construction of even degree polynomial Let $a=i+\sqrt 2$ ,Construct a polynomial $f (x)$ with integer coefficients such that $f  (a)=0$
I am getting that  it must be a degree polynomial with degree greater than 2
It is not of degree $2 $
But how will I construct
 A: Let
$$a = i + \sqrt{2}$$
Rearrange as
$$a - i = \sqrt{2}$$
Now, take the square
$$(a-i)^2 = (\sqrt{2})^2$$
$$a^2-2ia+(-1)=2$$
$$a^2-3=i\cdot 2a$$
Take the square again
$$(a^2-3)^2=(i\cdot 2a)^2$$
$$a^4-6a^2+9=-4a^2$$
$$a^4-2a^2+9=0$$
Note that this equation has four solutions, namely
$$\pm i \pm \sqrt{2}$$
A: Since $f(x)$ is a polynomial with integer coefficients, you know that $\sqrt2-i$ is also a root. So far you have $$(x-\sqrt2-i)(x-\sqrt2+i)=x^2-2\sqrt2x+3$$which doesn't yet work. 
Also if $f(x)$ is a polynomial with integer coefficients, then since the roots arise from something of the form "$\pm\sqrt\,$", you have that $i-\sqrt2$ is also a root, and thus so is $-i-\sqrt2$. This would give the polynomial $$(x+\sqrt2-i)(x+\sqrt2+i)=x^2+2\sqrt2x+3$$
Multiplying the two polynomials above by each other, $$(x^2-2\sqrt2x+3)(x^2+2\sqrt2x+3)=x^4-2x^2+9$$
You can check this by computing $(\sqrt2+i)^n$ for $n=2,4$. This gives $1+2\sqrt2i$ and $-7+4\sqrt2i$. So $$\begin{align}f(\sqrt2+i)&=(\sqrt2+i)^4-2(\sqrt2+i)^2+9\\&=-7+4\sqrt2i-2-4\sqrt2i+9\\&=0\end{align}$$
