Problem 46
Question
In Exercises \(45-50,\) perform the indicated operation(s) and write the result in standard form. $$ (8+9 i)(2-i)-(1-i)(1+i) $$
Step-by-Step Solution
Verified Answer
Therefore, the solution in the standard form for the given expression \( (8+9 i)(2-i)-(1-i)(1+i) \) is \(25 + 10i\).
1Step 1: Distribute the Multiplication
The first step is to distribute in both the expressions: \( (8 + 9i)(2 - i) \) equates to \(16 - 8i + 18i - 9i^2\),and \( (1 - i)(1 + i) \) equates to \(1 + i - i - i^2\).
2Step 2: Simplify Expressions
Simplify both expressions:For \(16 - 8i + 18i - 9i^2\), combine like terms and change \(i^2\) to \(-1\) (as \(i^2 = -1\)), obtaining: \(16 + 10i + 9 = 25 + 10i\).For \(1 + i - i - i^2\), again combine like terms and change \(i^2\) to \(-1\), obtaining: \(1 - 1 = 0\).
3Step 3: Subtract the Simplified Expressions
Now subtract \(25 + 10i\) and \(0\) as per the original expression, giving us the solution \(25 + 10i - 0 = 25 + 10i\).
Key Concepts
Distributive PropertySimplifying ExpressionsImaginary Unit
Distributive Property
The distributive property is a foundational algebra concept used to simplify expressions where $$ one term is multiplied by two or more terms inside parentheses. It states:
- For real numbers, if you have an expression like \(a(b + c)\), it becomes \(ab + ac\).
- \(8 \cdot 2 = 16\)
- \(8 \cdot (-i) = -8i\)
- \(9i \cdot 2 = 18i\)
- \(9i \cdot (-i) = -9i^2\)
Simplifying Expressions
Simplifying expressions, especially those with complex numbers, involves combining like terms and
$
$ substituting known values.
- First, recognize that $i^2 = -1$ for all calculations.
- $-8i + 18i = 10i$
- This transforms the expression to $16 + 10i + 9$.
- $1 - i + i - i^2$ becomes $1 - (-1) = 2$.
Imaginary Unit
The imaginary unit \(i\) is a mathematical concept used to extend real numbers into the complex plane. It$$ is defined by the property \(i^2 = -1\). Here's why it's crucial:
- Imaginary numbers arise naturally when solving equations like \(x^2 + 1 = 0\).
- The solution \(x = \pm i\) expands mathematics beyond the reals.
- This substitution transforms complex arithmetic into something more straightforward by leveragingknown values.
Other exercises in this chapter
Problem 45
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