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Chapter 2: Problem 39

Perform the operation and write the result in standard form. \((2+3 i)^{2}+(2-3 i)^{2}\)

Short Answer

Expert verified
The result of the operation \((2+3 i)^{2}+(2-3 i)^{2}\) in standard form is -10.

Step by step solution

01

Square the Complex Numbers

First, square each of the complex numbers. When squaring a binomial, use the formula \((a+b)^2 = a^2 + 2ab + b^2\). Hence,\((2+3i)^2 = (2)^2 + 2*2*3i + (3i)^2 = 4 + 12i - 9 = -5 + 12i\)\((2-3i)^2 = (2)^2 - 2*2*3i + (3i)^2 = 4 - 12i - 9 = -5 - 12i\)
02

Add the Squared Complex Numbers

Next, add the squared complex numbers. When adding complex numbers, you simply add the corresponding real parts and imaginary parts:\(-5 + 12i + (-5 - 12i) = -5 -5 + 12i -12i = -10 + 0i = -10\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Theorem
The Binomial Theorem is a handy tool that helps us expand expressions raised to a power, like \(a + b\)^n\. It especially comes in handy when dealing with quadratics or higher-degree polynomials. For a basic square, the formula \(a + b\)^2 = a^2 + 2ab + b^2\ is crucial. In this exercise, each complex number \(2 + 3i\) and \(2 - 3i\) is a binomial.
  • First, square the first term: \(a^2\).
  • Next, double the product of the two terms: \(+2ab\).
  • Finally, add the square of the second term: \(b^2\).
For \(2 + 3i\)^2\, you get: \((2)^2 + 2(2)(3i) + (3i)^2 = 4 + 12i - 9\).

Similarly, for \(2 - 3i\)^2\, it becomes: \((2)^2 - 2(2)(3i) + (3i)^2 = 4 - 12i - 9\). Using the theorem makes squaring more straightforward.
Complex Arithmetic
In complex arithmetic, we deal with numbers composed of a real part and an imaginary part. A complex number is typically written in the form \(a + bi\), where \(a\) is the real part and \(bi\) is the imaginary part. Understanding how to perform arithmetic operations with complex numbers is essential.
  • When squaring, derive real results from real-real and imaginary-imaginary interactions, while mixing (real-imaginary) contributes to the imaginary part.
  • Addition involves combining similar parts: real parts together, imaginary parts together.
In this task, after squaring \(2 + 3i\) and \(2 - 3i\), the resulting expressions are added: \(-5 + 12i\) and \(-5 - 12i\).

Adding these two results, \(-5\) and \(-5\) combine for \(-10\), while \(12i\) and \(-12i\) cancel each other out, resulting in a purely real sum of \(-10\).
Imaginary Unit
The imaginary unit \(i\) is a fundamental component of complex numbers and represents the square root of \(-1\). Its unique property, \(i^2 = -1\), helps us navigate complex arithmetic. In this exercise, \(3i\) consistently appears in calculations involving \(i\).
  • Squaring the imaginary component results in a real number: \((3i)^2 = 9i^2 = -9\).
  • This transformation demonstrates the power of \(i\) altered through exponents.
Comprehending this property enables correct handling of complex numbers, ensuring accuracy in results.

Thus, understanding the imaginary unit is essential for solving any problem that involves complex numbers, especially when calculating squares or performing operations as in the given exercise.

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Most popular questions from this chapter

Use a graphing utility to graph the equation. Use the graph to approximate the values of \(x\) that satisfy each inequality. \(y=-x^{2}+2 x+3 \quad\) (a) \(y \leq 0 \quad\) (b) \(y \geq 3\)

Find the key numbers of the expression. \(\frac{x}{x+2}-\frac{2}{x-1}\)

Determine whether each value of \(x\) is a solution of the inequality. Inequality. \(\begin{array}{lll}\frac{x+2}{x-4} \geq 3 & \text { (a) } x=5 & \text { (b) } x=4 \\ & \text { (c) } x=-\frac{9}{2} & \text { (d) } x=\frac{9}{2}\end{array}\)

Solve the inequality and graph the solution on the real number line. \(x^{2}-6 x+9<16\)

Solve the inequality and write the solution set in interval notation. \(2 x^{3}-x^{4} \leq 0\)
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