Math Problem Statement

Multiply the complex numbers (-6 - 8i)(3 + 4i)

Solution

It seems like I can’t do more advanced data analysis right now. Please try again later. However, I can still help you solve the complex multiplication from the image. Let's break it down manually.

You need to multiply the complex numbers (68i)(-6 - 8i) and (3+4i)(3 + 4i).

To multiply two complex numbers, we apply the distributive property (similar to multiplying two binomials) and simplify:

(68i)(3+4i)=(6)(3)+(6)(4i)+(8i)(3)+(8i)(4i)(-6 - 8i)(3 + 4i) = (-6)(3) + (-6)(4i) + (-8i)(3) + (-8i)(4i)

Now let's calculate each part:

  1. (6)(3)=18(-6)(3) = -18
  2. (6)(4i)=24i(-6)(4i) = -24i
  3. (8i)(3)=24i(-8i)(3) = -24i
  4. (8i)(4i)=32i2(-8i)(4i) = -32i^2, and since i2=1i^2 = -1, this becomes 32(1)=32-32(-1) = 32

Now, combine all the terms: 1824i24i+32-18 - 24i - 24i + 32 Simplify: (18+32)+(24i24i)=1448i(-18 + 32) + (-24i - 24i) = 14 - 48i

So, the product of the two complex numbers is:

1448i14 - 48i

Let me know if you need further explanation or have any questions!

Here are 5 related questions:

  1. How do you add and subtract complex numbers?
  2. How do you find the modulus of a complex number?
  3. What is the conjugate of a complex number, and how is it used?
  4. How do you divide complex numbers?
  5. How are complex numbers represented graphically?

Tip: Always remember that i2=1i^2 = -1 when multiplying complex numbers!

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Math Problem Analysis

Mathematical Concepts

Complex Numbers
Multiplication of Complex Numbers
Distributive Property

Formulas

Multiplication of complex numbers: (a + bi)(c + di) = (ac - bd) + (ad + bc)i
i^2 = -1

Theorems

Distributive Property

Suitable Grade Level

Grades 10-12