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Chapter 8: Problem 47

Find the following products. \((3+2 i)^{2}\)

Short Answer

Expert verified
The product is \(5 + 12i\).

Step by step solution

01

Identify the Expression Form

Recognize that the expression provided is \((3 + 2i)^{2}\). This is a complex number raised to the power of 2.
02

Apply Binomial Theorem

Use the binomial theorem for the square of a binomial. In general, \((a + b)^{2} = a^{2} + 2ab + b^{2}\). Here, let \(a = 3\) and \(b = 2i\).
03

Calculate Each Part Separately

First calculate \(a^{2}\):\((3)^{2} = 9\).Then \(2ab\):\(2 \cdot 3 \cdot 2i = 12i\).Finally, calculate \(b^{2}\):\((2i)^{2} = 4i^{2} = 4(-1) = -4\).
04

Sum the Results

Combine all calculated parts:\(a^{2} + 2ab + b^{2} = 9 + 12i - 4\). Simplify by combining like terms:\(5 + 12i\).

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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 provides a way to expand expressions that are raised to a power. It is used frequently in algebra to expand binomials. This theorem states:
  • For any binomial expression of the form o(a + b)^{n}= \sum_{k=0}^{n} \binom{n}{k}a^{n-k}b^{k}.
  • For our specific exercise with o(a + b)^{2}, we use a simplified version: o(a + b)^{2} = a^{2} + 2ab + b^{2}.
  • Here, the terms of the expansion represent different parts: a^{2} yields the square of the first term, \(2ab\) accounts for two times the product of both terms, and b^{2} gives the square of the second term.
The binomial theorem is handy for finding results of powers without needing to multiply the binomial repeatedly. Understanding how to use it with complex numbers is crucial in simplifying expressions, just like in the exercise you're solving.
Imaginary Unit
The imaginary unit, denoted as \(i\), is a mathematical concept used to handle numbers that involve the square root of negative one. Normally, taking the square root of a negative number isn鈥檛 possible with real numbers. But with complex numbers, it becomes feasible using \(i\). Here's more about this pivotal concept:
  • Imaginary unit \(i\) is defined as \(i = \sqrt{-1}\) which implies \(i^{2} = -1\).
  • It allows extension of the real number system to include complex numbers, which have the form \(a + bi\).
  • In our exercise, when we calculate \((2i)^{2}\), it's vital to remember \(i^{2} = -1\) so \((2i)^{2} = 4i^{2} = 4(-1) = -4\).
With \(i\), we can carry out operations, including addition, multiplication, and exponentiation, with numbers previously inaccessible due to their negative square roots.
Complex Multiplication
Complex multiplication involves multiplying two complex numbers, taking both the real and imaginary parts into account. Here's how to handle it:
  • Consider two complex numbers: \((a + bi)\) and \((c + di)\).
  • The product of these would be: \((a + bi)(c + di) = ac + adi + bci + bdi^{2}\).
By recognizing key steps:
  • Contribute each part: multiply the real parts to get one component, real and imaginary parts for another, consider \(i^{2}=-1\) for another.
  • Sum up these separate products, simplifying based on properties of \(i\).
  • In our example, \((3+2i)^{2}\) represents a self-product of a complex number; applying simplified binomial logic helps find \(5 + 12i\) as combined parts.
Understanding complex multiplication aids in working smoothly with expressions like the one in your task, simplifying otherwise daunting computations.

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Most popular questions from this chapter

Use a calculator to help write each complex number in standard form. Round the numbers in your answers to the nearest hundredth.\(1 \operatorname{cis} 261^{\circ}\)

If \(z\) is a complex number, show that the product of \(z\) and its conjugate is a real number.

Write each complex number in trigonometric form. Round all angles to the nearest hundredth of a degree.\(3-4 i\)

Write each complex number in trigonometric form. Round all angles to the nearest hundredth of a degree.\(-11+2 i\)

Is subtraction with complex numbers a commutative operation?
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