91Ó°ÊÓ

The Binomial Theorem is a powerful tool in algebra. It's used to expand expressions that are raised to a power, like \((a+b)^n\). This method is particularly useful when dealing with polynomial expressions.
When expanding simple binomial expressions such as \((1+i)^2\), we apply the binomial identity:

• \((a+b)^2 = a^2 + 2ab + b^2\)

Here, \(a=1\) and \(b=i\). So, replacing those in the formula, you expand \((1+i)^2\) to get:

• \(1^2 + 2 \times 1 \times i + i^2 = 1 + 2i + i^2\)

This outcome makes use of the identity of the imaginary unit, where \(i^2\) becomes \(-1\). Thus, after substituting \(i^2 = -1\), we end up with \(1 + 2i - 1\), which simplifies further to \(2i\).
The Binomial Theorem doesn't stop there; it extends to higher powers as well, like \((1-i)^3\). Using the theorem again, we use:

• \(a^3 + 3a^2b + 3ab^2 + b^3\)

Given \(a=1\) and \(b=-i\), this expands to \(1 - 3i + 3(-1) + i = -2 - 2i\) after substituting the known powers of \(i\).

The imaginary unit, represented by \(i\), is a fundamental concept in complex numbers. It is defined such that \(i^2 = -1\). This characteristic allows us to work with numbers that have no real solution.
When dealing with powers of \(i\), it cycles through values as follows:

• \(i^1 = i\)
• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\)

and then repeats. This cyclical nature is used to simplify complex expressions. In the problem, we see the expansion of binomials using this property, like \(i^2 = -1\), easily simplifying terms from expressions. For instance, the term \(2i^2\) becomes \(-2\).
Recognizing this cycle allows you to efficiently handle complex calculations, providing a clearer path to solutions.

Complex multiplication involves multiplying two complex numbers together. Each complex number is in the form \(a + ib\), with a real and an imaginary part.
In practice, this involves using the distributive property (also known as the FOIL method for binomials):

• When you multiply \(c+di\) and \(e+fi\), it means computing: \(c \, e + c \, fi + di \, e + di \, fi\)

After expansion, you apply simplification rules of \(i\) (where \(i^2 = -1\)) to simplify to a form \(x + yi\).
In the exercise, we multiply the results \(2i\) and \(-2 - 2i\) together. Start by multiplying like this:

• \(-4i = 2i(-2)\)
• \(4i^2 = 2i(-2i)\)

This implies \(-4i + 4(-1)\), simplifying finally to \(-4 - 4i\).
The result, \(-4 - 4i\), is a complex number with components clearly separated into real part \(-4\) and imaginary part \(-4\). These processes show how multiplication of complex numbers combines algebraic rules and the properties of \(i\).