Chapter 8: Problem 69
$$ \text { For Exercises 65-76, simplify and express in standard form. } $$ $$ (5-2 i)^{2} $$
Short Answer
Expert verified
The simplified expression in standard form is \(21 - 20i\).
Step by step solution
01
Set up the expression
We are asked to simplify the expression \((5-2i)^2\). To start, we write it as the square of a binomial: \((5 - 2i)(5 - 2i)\).
02
Apply the distributive property
Now, apply the distributive property to expand \((5 - 2i)(5 - 2i)\). First, multiply: \((5)(5) = 25\) \((5)(-2i) = -10i\) \((-2i)(5) = -10i\) \((-2i)(-2i) = 4i^2\) (Remember that \(i^2 = -1\); hence \(4i^2\) becomes \(4(-1) = -4\)).
03
Combine like terms
Now combine all the terms from the expansion: \[25 - 10i - 10i - 4\] First, add the real components: \(25 - 4 = 21\). Next, combine the imaginary components: \(-10i - 10i = -20i\).
04
Write the result in standard form
The standard form for a complex number is \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. So in this expression, the standard form is: \[21 - 20i\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Binomial Expansion
When dealing with expressions like \((a - b)^2\), we talk about binomial expansion. A binomial is simply an expression involving two terms, and the expansion refers to multiplying it out fully. In our exercise, we start with the binomial \((5 - 2i)^2\). Instead of calculating directly, we rewrite the expression as \((5 - 2i)(5 - 2i)\), which makes it easier to expand using multiplication.
- The first step is to apply the expression twice, based on the binomial theorem, such as \((a - b)(a - b)\), and calculate it as \(a^2 - 2ab + b^2\).
- This helps in breaking down the expression into smaller, more manageable calculations.
Distributive Property
A vital tool in multiplication within algebra, the distributive property helps to simplify expressions. It's defined as \(a(b + c) = ab + ac\) and applies to multiplication across terms within parentheses. In our example, we use it to expand \((5 - 2i)^2\).
- The distributive property enables multiplying each term inside the parentheses separately.
- This means calculating \(5 \cdot 5, 5 \cdot (-2i), (-2i) \cdot 5,\) and \((-2i) \cdot (-2i)\).
- It systematically breaks down complex multiplications into simpler parts, leading us to a complete expansion.
Imaginary Unit
The imaginary unit \(i\) forms the basis of complex numbers. Defined by the property \(i^2 = -1\), this allows us to handle square roots of negative numbers. In our scenario, when expanding \((5 - 2i)^2\), we encounter the term \((-2i) \cdot (-2i)\).
- Normally, \((-2)(-2)= 4\), but because of the presence of \(i\), this becomes \(4i^2\).
- \(i^2\)'s property transforms \(4i^2\) into \(4(-1)\), simplifying to \(-4\).
- This conversion is crucial to obtaining the correct real number during multiplication.