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(15-4i)(6-3i)=a+bi

2 min read Jun 16, 2024
(15-4i)(6-3i)=a+bi

Solving Complex Number Multiplication: (15-4i)(6-3i) = a + bi

This article will guide you through the steps of multiplying two complex numbers and expressing the result in the standard form of a complex number (a + bi).

Understanding Complex Numbers

A complex number is a number that can be expressed in the form a + bi, where:

  • a and b are real numbers
  • i is the imaginary unit, defined as the square root of -1 (i² = -1)

Multiplication Process

To multiply complex numbers, we use the distributive property (FOIL method) just like we do with binomials.

  1. Expand the product: (15 - 4i)(6 - 3i) = (15 * 6) + (15 * -3i) + (-4i * 6) + (-4i * -3i)

  2. Simplify: 90 - 45i - 24i + 12i²

  3. Substitute i² with -1: 90 - 45i - 24i + 12(-1)

  4. Combine real and imaginary terms: (90 - 12) + (-45 - 24)i

  5. Simplify further: 78 - 69i

Final Answer

Therefore, (15 - 4i)(6 - 3i) = 78 - 69i, where a = 78 and b = -69.

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