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Chapter 7: Problem 96

Perform the operation and write the result in standard form. $$-2 i(1+6 i)$$

Short Answer

Expert verified
The simplified result in standard form is \(12 - 2i\).

Step by step solution

01

Distribute \( -2i \) across \( (1 + 6i) \)

Apply the distributive property, which states that \(a(b + c) = ab + ac\), on \( -2i(1 + 6i) \). This gives us \(-2i * 1 - 2i * 6i \)
02

Simplify Multiplication

Simplify the multiplication to get \(-2i - 12i^2\). Remember that \(i^2 = -1\).
03

Substitute \(i^2\) with \(-1\)

Substitute \(i^2\) with \(-1\) to get \(-2i - 12(-1)\) which simplifies to \(-2i + 12\).
04

Write in standard form

The standard form of a complex number is \(a + bi\). So, reposition the terms to get this in the standard form as \(12 - 2i\).

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Key Concepts

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

Distributive Property
The distributive property is a fundamental mathematical principle used to multiply a single term across a sum or difference. It can be described using the formula:
  • For any numbers or expressions, if you have \(a(b+c)\), apply the operation as \(ab + ac\).
This principle helps to simplify expressions, particularly in algebra. In the context of complex numbers such as \(-2i(1 + 6i)\):
  • We can see \(-2i\) as the factor that needs to be distributed across each term inside the parentheses.
  • We perform the operation as: \(-2i \times 1 + (-2i) \times (6i)\).
Applying the distributive property is crucial in this exercise to separate the imaginary unit \(i\) and correctly solve for each part individually.
Simplifying Multiplication
Once you have used the distributive property, the next step is simplifying the multiplication. Simplifying involves performing straightforward multiplication tasks but with the added component of imaginary numbers. Consider these steps:
  • First, compute each part separately, such as \(-2i \times 1 = -2i\).
  • Then handle the more complex part, \(-2i \times 6i = -12i^2\).
Here it's important to remember that \(i^2\) equals \(-1\), a property that significantly changes results involving \(i\). So when multiplying with \(i^2\), like in \(-12i^2\), it transforms into \(-12 \times -1 = 12\), simplifying the expression to \(12\).
Multiplication in complex numbers often requires this substitution, enabling a conversion from more abstract to routine calculations.
Complex Number Standard Form
Complex numbers have a standard form that makes them easier to work with: \(a + bi\), where \(a\) and \(b\) are real numbers. This standard form distinctly separates the real part from the imaginary part.
The result from applying the distributive property and simplifying multiplication in this exercise was \(-2i + 12\).
  • Rearranging it to align with the standard form means writing it as \(12 - 2i\), where \(12\) is the real part and \(-2i\) is the imaginary part.
Organizing complex numbers into their standard form helps in identifying real and imaginary components clearly, facilitating easier comparative and operational mathematics. This arrangement is foundational in complex number computations and is often the final required step in solving these problems.

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Most popular questions from this chapter

Find \(\mathbf{u} \cdot \mathbf{v},\) where \(\theta\) is the angle between \(\mathbf{u}\) and \(\mathbf{v}.\) $$\|\mathbf{u}\|=100,\|\mathbf{v}\|=250, \theta=\frac{\pi}{6}$$

Determine whether u and v are orthogonal, parallel, or neither. $$\begin{aligned} &\mathbf{u}=-\frac{9}{10} \mathbf{i}+3 \mathbf{j}\\\ &\mathbf{v}=-5 \mathbf{i}+\frac{3}{2} \mathbf{j} \end{aligned}$$

Use DeMoivre's Theorem to find the indicated power of the complex number. Write the result in standard form. $$(3-2 i)^{5}$$

Use DeMoivre's Theorem to find the indicated power of the complex number. Write the result in standard form. $$\left[3\left(\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}\right)\right]^{2}$$

Find two vectors in opposite directions that are orthogonal to the vector \(\mathbf{u}.\) (There are many correct answers.) $$\mathbf{u}=-\frac{5}{2} \mathbf{i}-3 \mathbf{j}$$
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