91Ó°ÊÓ

Complex conjugates are pairs of complex numbers with identical real parts but opposite imaginary parts. Given any complex number in the form of \(a + bi\), its conjugate will be \(a - bi\).
For instance, the complex solutions \(9i\) and \(-9i\) are examples of complex conjugates since both have a real part equal to zero and imaginary parts that are opposites.
Why are complex conjugates important? They are usually present in polynomial equations with real coefficients because they ensure that the imaginary components cancel each other when these equations are expanded.
For example, when you expand \((x - 9i)(x + 9i)\) to form the polynomial, the imaginary parts (\(-9i\) and \(+9i\)) cancel out, resulting in a polynomial with only real coefficients.
This is a crucial property for constructing polynomials with integer coefficients.

An important aspect of polynomial equations is their coefficients. Coefficients are the numbers that multiply the variables or powers of variables in the equation. In our exercise, we need these coefficients to be integers.
Why integer coefficients? Integers are whole numbers (both positive and negative, including zero) without any fractional or decimal components. For many practical and theoretical purposes, polynomials with integer coefficients offer simplicity and elegance.
As shown in the step-by-step solution:

• We start with our complex solutions \(9i\) and \(-9i\).
• We then form the factors \((x - 9i)\) and \((x + 9i)\).
• When expanded, this leads to the polynomial equation \(x^2 + 81 = 0\).

Notice, the coefficients in the final equation \(x^2 + 81 = 0\) are 1 for \(x^2\) and 81 - both are integers. This is a direct result of using complex conjugate pairs in forming the polynomial.

The difference of squares is a useful algebraic identity that simplifies the multiplication of conjugate pairs. The formula can be stated as:
\(a^2 - b^2 = (a - b)(a + b)\).
In the context of our exercise:

• We have two factors: \((x - 9i)\) and \((x + 9i)\).
• Applying the difference of squares formula: \((x - 9i)(x + 9i) = x^2 - (9i)^2\).
• Simplify further: Since \(i^2 = -1\), \((9i)^2 = 81 \times -1 = -81\).
• Therefore, the expression becomes \(x^2 - (-81)\) or \(x^2 + 81\).

The result is an equation \(x^2 + 81 = 0\), neatly expanded using the difference of squares identity.
This is a powerful technique to transform products of binomials, especially conjugate pairs, into simpler polynomial forms with integer coefficients.