Find the polynomial with lowest degree having the following zeros : 2 and 2i .
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By definition, all zeroes of a polynomial are, in fact, the roots of that polynomial. According to the rule, the complex solutions come in complex conjugate pairs, therefore if 2i is a solution of the equation, it's conjugate -2i is also the solution of the equation.
We'll start from the fact that a polynomial could be written as a product of linear factors, if we know it's zeroes.
x1 = 2
x2 = 2i
x3 = -2i
The polynomial is:
P(x) = a(x - x1)(x - x2)(x - x3)
We'll substitute x1,x2 and x3:
P(x) = (x - 2)(x - 2i)(x + 2i)
We'll write the product (x - 2i)(x + 2i) as a difference of squares:
(x - 2i)(x + 2i) = x^2 + 4
P(x) = (x - 2)(x^2 + 4)
We'll remove the brackets:
P(x) = x^3 + 4x - 2x^2 - 8
The lowest degree polynomial, having as roots the values 2, 2i, -2i, is: P(x) = x^3 - 2x^2 + 4x - 8.
A polynomial has complex roots in pairs of conjugates. As the polynomial has roots 2 and 2i, it also has -2i as a root.
The polynomial is: (x - 2)(x - 2i)(x + 2i)
=> (x - 2)(x^2 - 4i^2)
=> (x - 2)(x^2 + 4)
=> x^3 - 2x^2 + 4x - 8
The required polynomial is x^3 - 2x^2 + 4x - 8
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