Given a=square root of 3 - i and f(x) = x^4-4x^2+16, prove that f(a)=0.

If f(a)=0, then a = sqrt3 - i is the root of the
polynomial f(x).

We'll substitute x by a and we'll verify
if f(a) = 0

f(a) = a^4 - 4a^2 +
16

f(a) = a^2(a^2 - 4) +
16

We'll re-write the difference of squares a^2 - 4 =
(a-2)(a+2)

a = sqrt3 - i

We'll
square raise both sides:

a^2 = (sqrt3 -
i)^2

We'll expand the
square:

a^2 = 3 - 2isqrt3 + i^2, where i^2 =
-1

a^2 = 2 - 2isqrt3

f(sqrt3 -
i) = (2 - 2isqrt3)(2 - 2isqrt3 - 4) + 16

We'll combine like
terms inside brackets:

f(sqrt3 - i) = (2 - 2isqrt3)(-2 -
2isqrt3) + 16

f(sqrt3 - i) = -(2 - 2isqrt3)(2 + 2isqrt3) +
16

We'll write the product as a difference of
squares:

f(sqrt3 - i) = -(2^2 - 4*3*i^2) +
16

f(sqrt3 - i) = - (4 + 12) +
16

f(sqrt3 - i) = -16 +
16

f(sqrt3 - i) =
0

So, f(a) = 0, where a =  sqrt3 - i,
q.e.d.