Problem 70 - A Graphical Approach to Precalculus with Limits

Question

Multiply as indicated. Write each product in standand form. $$-5 i(4-3 i)^{2}$$

Step-by-Step Solution

Answer

The product in standard form is \(120 - 35i\).

Step 1: Expand the Square

Start by expanding the expression \((4 - 3i)^2\). Use the formula for the square of a binomial: \((a - b)^2 = a^2 - 2ab + b^2\). Substitute \(a = 4\) and \(b = 3i\): \[(4 - 3i)^2 = 4^2 - 2(4)(3i) + (3i)^2 \] \[= 16 - 24i + 9i^2\] Since \(i^2 = -1\), replace \(9i^2\) with \(-9\):\[16 - 24i - 9\] This simplifies to \(7 - 24i\).

Step 2: Multiply by \(-5i\)

Next, multiply \(-5i\) by the expanded form \(7 - 24i\). Distribute \(-5i\) to each term inside the parentheses:\[-5i(7 - 24i) = -5i \cdot 7 - 5i \cdot 24i\]Calculate each term:\[-5i \cdot 7 = -35i\]\[-5i \cdot 24i = -120i^2\]Again, use \(i^2 = -1\) to simplify \(-120i^2\) to \(120\):\[-35i + 120\] This can be rewritten in standard form as \(120 - 35i\).

Key Concepts

Binomial Expansion

The binomial expansion is a method for expanding expressions that are raised to a power. In our exercise, we start with the expression \((4 - 3i)^2\). To expand a binomial like this, we use the formula:

• \((a - b)^2 = a^2 - 2ab + b^2\)

Here, \(a = 4\) and \(b = 3i\). Applying the formula step-by-step is crucial in making sure every part of the expression is handled correctly. Consider:

• First, compute \(4^2\), which is 16.
• Next, find \(-2 \times 4 \times 3i\), which gives \(-24i\).
• Then, calculate \((3i)^2\). Since \(i^2 = -1\), this becomes \(9 \times -1 = -9\).

Bringing it all together, the expanded form of \((4 - 3i)^2\) is \(16 - 24i - 9\). Simplifying this expression results in \(7 - 24i\). Binomial expansion is very helpful in algebra, especially when dealing with complex numbers.

Imaginary Unit

The imaginary unit, denoted as \(i\), is a crucial component in complex numbers. It is defined by its unique property:

• \(i^2 = -1\)

In solving the exercise, we encounter \((3i)^2\) while performing the binomial expansion. This requires us to use the property \(i^2 = -1\) to convert \(9i^2\) to \(-9\). Understanding how the imaginary unit operates simplifies complex arithmetic significantly. When multiplying \(-5i\) by \(24i\) in the next step, we get \(-120i^2\). Recognizing that \(i^2 = -1\) transforms this into \(120\). The imaginary unit allows complex numbers to be manipulated similarly to real numbers, making it a powerful tool for solving equations and performing calculations involving both real and imaginary components.

Standard Form of Complex Numbers

Complex numbers are typically expressed in standard form, which is written as \(a + bi\), where \(a\) and \(b\) are real numbers. This form clearly separates the real part, \(a\), from the imaginary part, \(bi\). Our final product from the exercise is \(120 - 35i\), which fits the standard form perfectly:

• The real part is \(120\).
• The imaginary part is \(-35i\).

This format makes it easier to understand and work with complex numbers in equations and other mathematical operations. The standard form is especially important in ensuring clarity in mathematics, as it helps distinguish between different components a complex number might have. By converting expressions into this useful form, students can perform operations and solve problems more efficiently and accurately. Whether adding, subtracting, or multiplying complex numbers, achieving the standard form ensures you are organized and precise.