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Chapter 10: Problem 80

Divide. Give answers in standard form. $$ \frac{-1+5 i}{3+2 i} $$

Short Answer

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Step by step solution

01

- Multiply by the conjugate

To divide complex numbers, multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of the denominator (3 + 2i) is (3 - 2i). So, multiply: Multiply: Multiply numerator: (-1 + 5i)(3 - 2i) = -1*3 - 1*(-2i) + 5i*3 + 5i*(-2i) = -3 + 2i + 15i - 10i^2 Replace i^2 with -1: -3 + 2i + 15i - 10(-1) = -3 + 2i + 15i + 10 = 7 + 17i Multiply denominator: (3 + 2i)(3 - 2i) = 3*3 + 3*(-2i) + 2i*3 - 2i*2i = 9 - 6i + 6i + 4i^2 Replace i^2 with -1: 9 - 6i + 6i + 4(-1) = 9 + 0 - 4 = 5
02

- Write final simplified form

Combine the results from Step 1: 7 = Unify numerator and denominator to get to solution: Simplified Example steps to the solution: Write final form: numerator/denominator = expression with i simplified Answer: write solution here

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Key Concepts

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

Complex Conjugate
To understand complex number division, we first need to know about the complex conjugate. The complex conjugate of a complex number is formed by changing the sign of its imaginary part.
So, the complex conjugate of \(3 + 2i\) is \(3 - 2i\).
This is valuable because multiplying a complex number by its conjugate helps in eliminating the imaginary part in the denominator, simplifying our calculations.
For example, in the problem \(\frac{-1+5i}{3+2i}\), we multiply both the numerator and the denominator by the complex conjugate of the denominator, \(3 - 2i\).
Simplifying Complex Numbers
After multiplying by the complex conjugate, we simplify both the numerator and the denominator.
This involves carrying out the multiplication and combining like terms.

For the numerator:
\((-1 + 5i)(3 - 2i)\):
  • Multiply: \(-1*3 = -3\)
  • Combine imaginary parts: \(-1*(-2i) = 2i\) and \(5i*3 = 15i\)
  • Combine: \(2i + 15i = 17i\)
  • Handle \(i^2\): \(5i*(-2i) = -10i^2\)
  • Since \(i^2 = -1\), simplified: \(-10i^2 = 10\)
  • Add real numbers: \(-3 + 10 = 7\)
    • So, the numerator becomes \(7 + 17i\).

For the denominator:
  • \((3 + 2i)(3 - 2i) = 9 - 4 = 5\)
Thus, the denominator becomes \(5\).
So, the fraction \(\frac{7 + 17i}{5}\) becomes \(\frac{7}{5} + \frac{17i}{5}\).
Standard Form
Standard form for complex numbers is \(a + bi\), where \(a\) and \(b\) are real numbers.
This form is straightforward and makes operations simpler.

For our example, the final simplified answer in standard form is:
\(\frac{-1 + 5i}{3 + 2i} = \frac{7 + 17i}{5} = 1.4 + 3.4i\).
Always remember to express your final answer with the real part first, followed by the imaginary part. This method ensures clarity and accuracy in your calculations.

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Most popular questions from this chapter

List all of the following sets to which each number belongs. A number may belong to more than one set. real numbers pure imaginary numbers nonreal complex numbers complex numbers $$ \sqrt{-49} $$

List all of the following sets to which each number belongs. A number may belong to more than one set. real numbers pure imaginary numbers nonreal complex numbers complex numbers $$ -7 i $$

Work each problem. Suppose someone claims that \(\sqrt[n]{a^{n}+b^{n}}\) must equal \(a+b,\) because when \(a=1\) and \(b=0,\) a true statement results: $$ \sqrt[n]{a^{n}+b^{n}}=\sqrt[n]{1^{n}+0^{n}}=\sqrt[n]{1^{n}}=1=1+0=a+b $$ Explain why this is faulty reasoning.

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