Problem 39 - College Algebra and Calculus: An Applied Approach

Question

Perform the indicated operation and write the result in standard form. $$(5+6 i)^{2}$$

Step-by-Step Solution

Answer

The result in standard form is \(-11 + 60i\).

Step 1: Identify the Terms

First, recognize the real part and the imaginary part of the complex number. In this case, the real part is 5 and the imaginary part is 6. Therefore, our goal is to square the number \(5 + 6i\). To achieve this, we have to multiply this complex number by itself.

Step 2: Apply the Mechanics of Squaring

Next, perform the multiplication operation. We use the binomial formula: \((a+b)^2 = a^2 + 2ab + b^2\), where \(a\) is the real part (5) and \(b\) the imaginary part (6). Applying the formula, we get \(25 + 2*5*6i + (6i)^2\).

Step 3: Simplify the Result

Now, simply Calculate the above expression which gives us \(25 + 60i - 36\) (since \(i^2\) is -1). Once simplified, this yields \(-11 + 60i\).

Key Concepts

Binomial Formula Squaring

The binomial formula describes how to square a binomial—a two-term expression—by expanding it into a trinomial. To square a binomial, which is of the form \(a + b)\), you follow the pattern \(a^2 + 2ab + b^2\). This is known as squaring a binomial and is crucial in complex number arithmetic.

For the complex number \(5 + 6i\), which combines real and imaginary parts, the process is the same. Here's how the squaring works step by step:

• First, square the real part: \(5^2 = 25\).
• Then, square the imaginary part: \(6i^2 = -36\), since \(i^2 = -1\).
• Finally, multiply the two terms together and double them: \(2 \times 5 \times 6i = 60i\).

Combine these and you get the expanded form: \(25 + 60i - 36\), which simplifies to \(\-11 + 60i\). It is essential to understand this expansion method, as it forms the basis for all polynomial expansions involving complex numbers.

Multiplying Complex Numbers

Complex numbers are multiplied in a similar fashion to binomials. A complex number is formed of a real part and an imaginary part (e.g., \(a + bi\)). When you multiply two complex numbers, you use the distributive property to multiply each term in the first complex number by each term in the second complex number and then combine like terms.

For example, if you were to multiply \(a + bi\) by itself (which is essentially squaring the number), the process would be the same as squaring a binomial:

• First, multiply the real parts: \(a \times a = a^2\).
• Then, multiply the imaginary parts, remembering \(i^2 = -1\): \(bi \times bi = -b^2\).
• Multiply the real part of the first number by the imaginary part of the second and double it, since you do it twice in the process: \(2 \times a \times bi\).

This process emphasizes the complex plane's rules, where the product of two imaginary numbers yields a real number because \(i^2 = -1\). Therefore, the multiplication of complex numbers requires careful execution of normal algebraic processes, but with additional attention to the special case of \(i^2\).

Imaginary Numbers Arithmetic

Imaginary numbers involve arithmetic slightly different from real numbers due to the presence of the imaginary unit \(i\), where \(i^2 = -1\). The fundamental operations—addition, subtraction, multiplication, and division—all follow similar rules as real numbers, with a unique property twist due to the \(i\) factor.

In the context of imaginary numbers:

• To add or subtract, simply combine like terms of the real parts and the imaginary parts separately, similar to combining like terms in algebraic expressions.
• Multiplication involves using the distributive property while remembering that \(i^2 = -1\), which will convert terms with \(i^2\) into real numbers.
• Division often requires the use of complex conjugates to eliminate the imaginary parts from the denominator, resulting in a real number.

Understanding these rules and the unique element of \(i\) is crucial for any type of arithmetic involving imaginary numbers, ensuring you can perform operations correctly and arrive at the proper standard form of a complex number as a result.