Problem 70
Question
Multiply as indicated. Write each product in standard form. $$-5 i(4-3 i)^{2}$$
Step-by-Step Solution
Verified Answer
The product is
\(-120 - 35i\).
1Step 1: Expand the Expression
First, we need to expand the expression \((4 - 3i)^2\). Apply the formula for squaring a binomial: \((a - b)^2 = a^2 - 2ab + b^2\). Here, \(a = 4\) and \(b = 3i\), so we calculate: \(4^2 - 2 \times 4 \times 3i + (3i)^2\).
2Step 2: Simplify the Squared Terms
Calculate each part: \(4^2 = 16\), \(-2 \times 4 \times 3i = -24i\), \((3i)^2 = 9i^2\). Since \(i^2 = -1\), \(9i^2 = 9(-1) = -9\). Now, combine all components: \(16 - 24i - 9\).
3Step 3: Combine Like Terms
Combine the real parts: \(16 - 9 = 7\). This gives us a simplified expression: \(7 - 24i\).
4Step 4: Distribute the Scalar Multiple
Multiply \(-5i\) by each term in the expression \((7 - 24i)\).Calculate: \(-5i \times 7 = -35i\) \(-5i \times (-24i) = 120i^2\). Since \(i^2 = -1\), substitute to get \(120i^2 = 120(-1) = -120\).
5Step 5: Final Solution
Combine the terms from the multiplication: \(-120 - 35i\). This is the product in standard form.
Key Concepts
Imaginary unitBinomial expansionComplex multiplicationStandard form of complex numbers
Imaginary unit
The imaginary unit, denoted as \(i\), is a fundamental concept in mathematics, particularly when dealing with complex numbers. The imaginary unit is defined by the property that \(i^2 = -1\). This gives rise to numbers that are called "imaginary" because they do not exist on the traditional number line. Instead, they exist on an "imaginary" number line perpendicular to the real number line:
- Imaginary numbers are written in the form \(bi\), where \(b\) is a real number.
- An example of an imaginary number is \(3i\), which equals \(3 \times i\).
Binomial expansion
The binomial expansion is a process used to expand expressions that are raised to a power. A binomial is a polynomial with two terms, such as \((a + b)\). When expanding a binomial like \((a - b)^2\), we use the formula:
- \((a - b)^2 = a^2 - 2ab + b^2\)
- First, calculate \(4^2 = 16\).
- Then, find \(-2 \times 4 \times 3i = -24i\).
- Finally, compute \((3i)^2 = 9i^2\). Since \(i^2 = -1\), this becomes \(-9\).
- \(16 - 24i - 9\) which simplifies to \(7 - 24i\).
Complex multiplication
Complex multiplication involves multiplying complex numbers, which can be in the form \((a + bi)\). The process is similar to the distributive property but includes handling \(i^2 = -1\).
- When you multiply two complex numbers, ensure each component is multiplied, including both the real and imaginary parts.
- An essential step is remembering that terms like \(i \times i = i^2\), which equals \(-1\).
- The term \(-5i\) is multiplied by \((7 - 24i)\).
- Calculate \(-5i \times 7\) to get \(-35i\).
- Then, \(-5i \times (-24i) = 120i^2\), which becomes \(-120\) after substituting \(i^2 = -1\).
Standard form of complex numbers
The standard form of complex numbers is expressed as \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit.
- The term \(a\) represents the real part of the complex number.
- The term \(bi\) represents the imaginary part.
- After multiplication and simplification, we acquire the result \(-120 - 35i\).
- Here, \(-120\) is the real part, and \(-35i\) is the imaginary part.
Other exercises in this chapter
Problem 70
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Use the given zero to completely factor \(P(x)\) into linear factors. $$\text { Zero: } i ; P(x)=x^{5}-x^{4}+5 x^{3}-5 x^{2}+4 x-4$$
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