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Chapter 8: Problem 55

Find the following products. \(3 i(1+i)^{2}\)

Short Answer

Expert verified
The product is -6.

Step by step solution

01

Expand the Square

First, expand the square in the expression. We need to expand \( (1+i)^{2} \): \((1+i)^2 = (1+i)(1+i) = 1 + 2i + i^2\). Since \( i^2 = -1 \), the expression simplifies to: \( 1 + 2i - 1 = 2i \). Therefore, \( (1+i)^2 = 2i \).
02

Distribute the Outer Term

Now, use the result from Step 1 to distribute \(3i\) in the expression \(3i imes (1+i)^2\): \(3i imes 2i = 3i imes 2i = 6i^2\).
03

Simplify the Expression

Since \( i^2 = -1 \), substitute \( -1 \) for \( i^2 \): \(6i^2 = 6 imes -1 = -6\). Thus, the product is \(-6\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Imaginary Unit
The imaginary unit, often denoted as \( i \), is a fundamental concept in complex numbers. It's essential to grasp this to handle complex numbers effectively.
  • \( i \) is defined by the relation \( i^2 = -1 \).
  • This means that \( i \) itself is the square root of \(-1\).
  • Using \( i \), a complex number can be expressed in the form \( a + bi \), where \( a \) and \( b \) are real numbers.
This definition contrasts with real numbers, where no real number squared can equal \(-1\). The imaginary unit allows us to extend calculus and algebra into this new realm, enabling the solving of equations that otherwise wouldn't have solutions in the real number system.
When working with expressions involving \( i \), remember that squaring \( i \) will always yield \(-1\), a vital simplification when performing operations like addition, subtraction, and multiplication within complex numbers.
The Process of Complex Multiplication
Complex multiplication involves a unique process, different from multiplying real numbers alone. Let's dive into what makes it distinct.
  • When multiplying two complex numbers, each part (the real and imaginary components) must be multiplied separately and then combined.
  • For example, consider \((a + bi)(c + di)\), the product is: \[ a \cdot c + a \cdot di + bi \cdot c + bi \cdot di = ac + adi + bci + bdi^2 \]
Notice how each term comes from distributing the individual parts of the complex numbers.
Because \( i^2 = -1 \), this simplifies further:
  • The term \( bdi^2 \) becomes \(-bd\), transforming the equation into \( (ac - bd) + (ad + bc)i \).
This outcome is a new complex number. Hence, multiplication in complex numbers combines both rotation and scaling in the complex plane, showcasing how this operation is richer than mere multiplication of real numbers.
Square of Complex Numbers
Squaring a complex number is a common operation, often involved in many algebraic simplifications. To square a complex number, follow these general steps:
  • Write the initial expression in the form \((a + bi)^2\).
  • Apply the identity: \[ (a + bi)^2 = (a + bi)(a + bi) = a^2 + 2abi + b^2i^2 \]
Here, by simplifying \( i^2 \) to \(-1\), the expression becomes:
  • \( a^2 - b^2 + 2abi \)
This illustrates how squaring introduces combined terms and a new imaginary component.
Squaring affects both the modulus and the argument of the complex number:
  • The modulus (or length) is squared: \( |a + bi|^2 = a^2 + b^2 \).
  • The argument (or angle) is doubled in the complex plane, altering the direction without changing the origin.
This process shows the powerful transformation properties of squaring, altering both the magnitude and direction of complex numbers in their plane.

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