91Ó°ÊÓ

The complex conjugate of a complex number is formed by changing the sign of its imaginary part. If you have a complex number in the form of \(a + bi\), its conjugate will be \(a - bi\).

Here’s an example: the complex number \(3 + 4i\) has a conjugate \(3 - 4i\).

When we multiply a complex number by its conjugate, an interesting thing happens. Let's see:

• \((a + bi)(a - bi) = a^2 - (bi)^2\)


• Since \(i^2 = -1\), the equation becomes \(a^2 + b^2\).

This means that the result is always a real number, as the imaginary parts cancel out.

In the given exercise, \(2+i\) and \(2-i\) are conjugates. By multiplying them, we get:
\((2+i)(2-i) = 2^2 - i^2 = 4 + 1 = 5\).
Notice, it's a real number! This helps simplify further calculations.

Multiplying complex numbers involves using the distributive property (also known as the FOIL method in algebra). To multiply two complex numbers like \((a+bi)(c+di)\), follow these steps:

• Multiply the real parts: \(a \times c\).


• Multiply the real part by the imaginary part: \(a \times di\).


• Multiply the imaginary part by the real part: \(bi \times c\).


• Multiply the imaginary parts: \(bi \times di = bdi^2\).

Combine these results and use \(i^2 = -1\) to simplify the expression:
\(ac + adi + bci + bdi^2 = ac + adi + bci - bd\).

Let’s apply this to our problem. We initially multiplied \(2+i\) with \(2-i\) to get \(5\), then we multiplied \(5\) by \(4+3i\):

• Multiply the real parts: \(5 \times 4 = 20\).


• Multiply the real part by the imaginary part: \(5 \times 3i = 15i\).

Combine the results:
\(20 + 15i\).
This is our product, and we will put it in standard form in the next section!

The standard form of complex numbers is expressed as \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit, which satisfies \(i^2 = -1\).

When dealing with complex numbers, it's crucial to ensure that your final answer is in this standard form. Here’s why:

• It's easier to identify the real part (\(a\)) and the imaginary part (\(b\)).


• It simplifies addition, subtraction, and comparison between complex numbers.

In our exercise, after multiplying \((2+i)(2-i)\) to get \(5\), and then multiplying \(5\) by \(4+3i\), we arrived at \(20 + 15i\).

This result is already in standard form, with \(20\) as the real part and \(15i\) as the imaginary part. Therefore, the final answer to the problem is
\(20 + 15i\).

Ensuring the result is in standard form helps you accurately describe and use complex numbers in further calculations and applications!