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Imaginary numbers are an essential part of complex number mathematics. They arise when we take square roots of negative numbers, a concept that doesn't exist in the realm of real numbers. The imaginary unit is denoted by the symbol \(i\), which is defined as \(i = \sqrt{-1}\). This means that when you square \(i\), you get \(-1\): \(i^2 = -1\).

Imaginary numbers are used in various fields, including engineering and physics, to solve problems involving square roots of negative quantities. They allow us to express numbers that wouldn't otherwise make sense, such as \(\sqrt{-4} = 2i\). When combining imaginary numbers with real numbers, we get complex numbers. For example, a number like \(3 + 2i\) is a complex number, where 3 is the real part and \(2i\) is the imaginary part.

Calculating the square roots of negative numbers is where imaginary numbers come into play. For instance, in the given exercise, we encounter \(\sqrt{-81}\). Normally, you cannot take a square root of a negative number because no real number multiplied by itself will yield a negative product.

This is where the imaginary unit \(i\) is used. For \(\sqrt{-81}\), you break it down as follows: \(\sqrt{-81} = \sqrt{-1 \times 81}\), which simplifies to \(\sqrt{-1} \times \sqrt{81}\). Since \(\sqrt{-1} = i\) and \(\sqrt{81} = 9\), you end up with \(9i\). This step transforms a seemingly impossible operation into one that is manageable using imaginary numbers.

• Always express \(\sqrt{-k}\) as \(i\sqrt{k}\) where \(k\) is a positive number.
• This approach simplifies many algebraic operations and equations.

In equations involving complex numbers, it's important to equate corresponding parts to solve for variables. Each complex number can be broken down into its real and imaginary components. For example, in the equation \(3a + 9i = 15 + 3bi\), identify the real and imaginary parts on each side separately.

• Real parts: \(3a\) and 15
• Imaginary parts: \(9i\) and \(3bi\)

By setting the real parts equal, you get \(3a = 15\). For the imaginary parts, \(9 = 3b\) is derived. Solving these gives the values for \(a\) and \(b\). Divide by the coefficients in front of \(a\) and \(b\) respectively to isolate them. You find \(a = 5\) and \(b = 3\).

This method is vital in solving any equation with complex numbers because it ensures that all components are balanced throughout the equation.