Math Problem Statement

Use a table to find the product. $(-3+0.6j)(0.4j-7)=$

Solution

To solve the expression (3+0.6j)(0.4j7)(-3 + 0.6j)(0.4j - 7), we can use the distributive property (also known as the FOIL method for binomials). Let's break it down step by step using a table for clarity.

We’ll set up the table for the distribution process:

Term3-30.6j0.6j
0.4j0.4j(3)(0.4j)=1.2j(-3)(0.4j) = -1.2j(0.6j)(0.4j)=0.24j2(0.6j)(0.4j) = 0.24j^2
7-7(3)(7)=21(-3)(-7) = 21(0.6j)(7)=4.2j(0.6j)(-7) = -4.2j

Now, let's compute the result:

  • The first row: 1.2j+0.24j2-1.2j + 0.24j^2
  • The second row: 214.2j21 - 4.2j

Putting it all together: (3+0.6j)(0.4j7)=214.2j1.2j+0.24j2(-3 + 0.6j)(0.4j - 7) = 21 - 4.2j - 1.2j + 0.24j^2 Simplify the imaginary terms: =215.4j+0.24j2= 21 - 5.4j + 0.24j^2

Since j2=1j^2 = -1, substitute 0.24j2=0.24(1)=0.240.24j^2 = 0.24(-1) = -0.24: =215.4j0.24= 21 - 5.4j - 0.24

Finally, combine the real terms: =20.765.4j= 20.76 - 5.4j

Thus, the product is: 20.765.4j\boxed{20.76 - 5.4j}

Let me know if you'd like more details or have any questions about the steps!

Here are some related questions to explore:

  1. How can the distributive property be applied to more complex expressions?
  2. What is the significance of j2=1j^2 = -1 in complex number operations?
  3. How do you handle higher powers of jj (like j3,j4j^3, j^4) in multiplication?
  4. What are other ways to visualize the multiplication of complex numbers?
  5. How do you solve equations involving complex numbers?

Tip: When multiplying complex numbers, always remember to use j2=1j^2 = -1 to simplify expressions with imaginary terms!

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Math Problem Analysis

Mathematical Concepts

Complex Numbers
Distributive Property
Imaginary Numbers

Formulas

(a + bj)(c + dj) = ac + adj + bcj + bdj^2
j^2 = -1

Theorems

Distributive Property
Complex Number Multiplication

Suitable Grade Level

Grades 9-12