Problem 45 - Algebra A Combined Function

Question

Multiply. Write your answers in the form \(a+b i\). $$ (4-2 i)^{2} $$

Step-by-Step Solution

Answer

The answer is \(12 - 16i\).

Step 1: Identify the Expression

We are given the complex expression \((4 - 2i)^2\) to simplify. This means we need to multiply the expression \((4 - 2i)\) by itself.

Step 2: Apply the Formula for Squaring a Binomial

The formula for squaring a binomial \((a - b)^2\) is \(a^2 - 2ab + b^2\). Identify \(a = 4\) and \(b = 2i\).

Step 3: Calculate \(a^2\)

Calculate \(a^2 = 4^2 = 16\).

Step 4: Calculate \(-2ab\)

Calculate \(-2ab = -2 \times 4 \times 2i = -16i\).

Step 5: Calculate \(b^2\)

Calculate \(b^2 = (2i)^2 = 4i^2\). Remember that \(i^2 = -1\), so \(4i^2 = 4 \times (-1) = -4\).

Step 6: Combine the Results

Substitute \(a^2\), \(-2ab\), and \(b^2\) from previous steps into the expression \(a^2 - 2ab + b^2\) to get: \(16 - 16i - 4\).

Step 7: Simplify the Expression

Combine the real parts and the imaginary parts: \((16 - 4) - 16i = 12 - 16i\).

Key Concepts

Multiplication of Complex Numbers

When dealing with complex numbers, we sometimes need to multiply them. A complex number is typically written in the form \(a+bi\), where \(a\) is the real part, and \(b\) is the imaginary part. Multiplying complex numbers requires using the distributive property, just like you would with binomials in algebra. In our exercise, \((4 - 2i)^2\), we want to multiply the number by itself.

• Expand the expression \((4 - 2i)(4 - 2i)\), ensuring each term in the first complex number is multiplied by each term in the second.
• Remember, when multiplying the imaginary parts, treat \(i\) as a variable, but keep in mind that \(i^2 = -1\).
• This method guarantees all parts of the numbers interact, forming a new complex number.

Multiplying complex numbers can look tricky, but with practice, you'll see that it's manageable.

Complex Conjugates

A complex conjugate flips the sign of the imaginary part of a complex number. For example, if our complex number is \(a+bi\), its complex conjugate is \(a-bi\).

• Complex conjugates are especially useful in division, and they help simplify expressions further.
• Using conjugates ensures any imaginary number terms cancel out when needed, by relying on the fact that \((a+bi)(a-bi) = a^2 + b^2\).

Although our exercise doesn't directly require using the conjugate, understanding this concept can provide more profound insight into manipulating complex numbers. Complex conjugates help in realizing the relationship between a number and its "mirror image" across the real axis.

Imaginary Unit i

The imaginary unit \(i\) is defined such that \(i^2=-1\). This property allows us to extend the real numbers into the complex plane.

• In our exercise, we see its impact when calculating \(b^2 = (2i)^2 = 4i^2\), where it immediately changes the result to a real number by replacing \(i^2\) with \(-1\).
• Another role \(i\) plays is in managing the results of operations involving complex numbers, keeping the structure \(a+bi\) intact while introducing real or imaginary components depending on the operation.

Understanding \(i\) is crucial, because it bridges the gap between real and complex number calculations, allowing new dimensions of problem-solving.

Binomial Expansion

The process of squaring a binomial involves the binomial expansion. It's crucial in our exercise, where \((4-2i)^2\) is expanded using the formula \((a-b)^2 = a^2 - 2ab + b^2\).

• The first step, \(a^2\), involves squaring \(4\) to get \(16\).
• The second step, \(-2ab\), accounts for the interaction between the two terms: \(-2 \times 4 \times 2i = -16i\).
• The third step, \(b^2\), includes squaring \(2i\) resulting in \(4i^2\), transforming into \(-4\) when recognizing that \(i^2 = -1\).
• All the parts are then combined leading to a simplified expression \(12 - 16i\).

Binomial expansion allows us to systematically calculate the result of powers of binomials, vital for sailing through solutions of complex expressions.