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

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As a is equal to sqrt 3 - i, we can substitute that for x in the expression f(x) = x^4 - 4x^2 + 16.

f(a) = [(sqrt 3 - i)]^4 - 4*[ (sqrt 3 - i)]^2 + 16

=> f(a) = (3 + i^2 - 2i*sqrt 3)^2 - 4*(3...

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As a is equal to sqrt 3 - i, we can substitute that for x in the expression f(x) = x^4 - 4x^2 + 16.

f(a) = [(sqrt 3 - i)]^4 - 4*[ (sqrt 3 - i)]^2 + 16

=> f(a) = (3 + i^2 - 2i*sqrt 3)^2 - 4*(3 + i^2 - 2i*sqrt 3) + 16

=> f(a) = (2 - 2i*sqrt 3)^2 - 4*(2 - 2i*sqrt 3) + 16

=> f(a) = 4(1 + (i*sqrt 3)^2 - 2*i*sqrt 3) - 8 + 8i*sqrt 3 + 16

=> f(a) = 4 + 4*i^2*3 - 8*i*sqrt 3 - 8 + 8i*sqrt 3 + 16

=> f(a) = 4 - 12  -8 + 16 -8*i*sqrt 3 + 8i*sqrt 3

=> f(a) = 0

Therefore for f(x) = x^4 - 4x^2 + 16, f(a) = 0 if a = sqrt 3 - i

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