91Ó°ÊÓ

The complex conjugate of a complex number is found by changing the sign of its imaginary part. For instance, if you have a complex number in the form of \(a + bi\), its conjugate will be \(a - bi\). The conjugate reflects the complex number over the real axis on the complex plane. In our exercise, the given complex number is 3-9i. To find its conjugate, we change the sign of the imaginary part, resulting in 3+9i.

Conjugates are significant in mathematics because multiplying a complex number by its conjugate results in a real number. Multiplying \(a + bi\) and \(a - bi\) gives us \(a^2 + b^2\). This property is useful in simplifying fractions and rationalizing denominators containing complex numbers.

The imaginary unit, denoted as \(i\), is defined as the square root of -1. This means that \(i^2 = -1\). The imaginary unit allows us to extend the real number system to the complex number system, which includes all numbers in the form \(a + bi\), where \(a\) and \(b\) are real numbers.

In the context of our exercise, the given complex number is expressed as 3-9i. Here, 3 is the real part and -9i is the imaginary part, where -9 is multiplied by the imaginary unit \(i\). Understanding \(i\) helps us perform operations such as addition, subtraction, and multiplication of complex numbers.

When multiplying two complex numbers, we follow the distributive property, similar to multiplying binomials. This involves expanding and then simplifying the expression. Let’s break down the multiplication of two complex numbers \( (a + bi)(c + di) \):

• Multiply the real parts: \( ac \)
• Multiply the real part of the first number by the imaginary part of the second: \( adi \)
• Multiply the imaginary part of the first number by the real part of the second: \( bci \)
• Multiply the imaginary parts: \( bdi^2 \) (since \( i^2 = -1 \), this becomes \( -bd \)).

Combining all these, we have: \[ ac + adi + bci - bd \] Simplifying further: \[ (ac - bd) + (ad + bc)i \]

For our exercise, we multiply \( 3 - 9i \) by its conjugate \( 3 + 9i \). Using the difference of squares formula: \[ (3 - 9i)(3 + 9i) = a^2 + b^2 \] When calculated: \[ 3^2 + (-9)^2 = 9 + 81 = 90 \] The final result is 90, a real number.