Math Problem Statement

Simplify (2+3i)⋅(1−4i)(2 + 3i) \cdot (1 - 4i)(2+3i)⋅(1−4i).

Solution

We are tasked with simplifying the expression (2+3i)(14i)(2 + 3i) \cdot (1 - 4i). Let's go step by step.

Step 1: Use the distributive property (FOIL method)

We distribute the terms: (2+3i)(14i)=2(1)+2(4i)+3i(1)+3i(4i)(2 + 3i)(1 - 4i) = 2(1) + 2(-4i) + 3i(1) + 3i(-4i) Now simplify each term:

  • 21=22 \cdot 1 = 2
  • 2(4i)=8i2 \cdot (-4i) = -8i
  • 3i1=3i3i \cdot 1 = 3i
  • 3i(4i)=12i23i \cdot (-4i) = -12i^2

Step 2: Simplify using i2=1i^2 = -1

Recall that i2=1i^2 = -1, so: 12i2=12(1)=12-12i^2 = -12(-1) = 12

Step 3: Combine all terms

Now substitute back into the expression: 28i+3i+122 - 8i + 3i + 12 Combine the real terms and the imaginary terms: (2+12)+(8i+3i)=145i(2 + 12) + (-8i + 3i) = 14 - 5i

Final Answer:

(2+3i)(14i)=145i(2 + 3i) \cdot (1 - 4i) = 14 - 5i

Would you like more details on this process or clarification on any step?

Here are 5 related questions you could explore:

  1. What happens when you multiply two complex numbers geometrically?
  2. How do you find the conjugate of a complex number?
  3. Can you simplify (3+2i)(2+5i)(3 + 2i)(2 + 5i)?
  4. What is the modulus of the complex number 145i14 - 5i?
  5. How do you divide two complex numbers?

Tip: When multiplying complex numbers, always remember that i2=1i^2 = -1, which simplifies the terms with imaginary units.

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Math Problem Analysis

Mathematical Concepts

Complex Numbers
Multiplication of Complex Numbers

Formulas

(a + bi)(c + di) = (ac - bd) + (ad + bc)i

Theorems

i^2 = -1

Suitable Grade Level

Grades 9-12