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Chapter 10: Problem 37

Write each quotient in the form \(a+b i\) $$ \frac{3+5 i}{1+i} $$

Short Answer

Expert verified
The quotient in the form \( a + bi \) is \( 4 + i \).

Step by step solution

01

Represent the Quotient

The given complex quotient is \( \frac{3+5i}{1+i} \). We need to write it in the form \( a + bi \).
02

Multiply by the Conjugate

Multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of \( 1+i \) is \( 1-i \). So, multiply by \( \frac{1-i}{1-i} \):\[\frac{(3+5i)(1-i)}{(1+i)(1-i)}\]
03

Simplify the Denominator

Calculate \( (1+i)(1-i) \) which is a difference of squares: \[ (1+i)(1-i) = 1^2 - (i)^2 = 1 - (-1) = 2 \] The denominator simplifies to 2.
04

Expand the Numerator

Distribute \( 3+5i \) and \( 1-i \) in the numerator: \[(3+5i)(1-i) = 3\cdot1 + 3\cdot(-i) + 5i\cdot1 + 5i\cdot(-i) = 3 - 3i + 5i - 5i^2\] Since \( i^2 = -1 \), it becomes:\[3 - 3i + 5i + 5 = 8 + 2i\] The expanded numerator is \( 8 + 2i \).
05

Write in the Form \( a + bi \)

Combine the results of the numerator and the denominator:\[\frac{8+2i}{2} = \frac{8}{2} + \frac{2i}{2} = 4 + i\] This is the quotient in the form \( a + bi \).

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Key Concepts

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

Complex Conjugate
In the realm of complex numbers, the complex conjugate of a given complex number is obtained by changing the sign of its imaginary part. For instance, the complex conjugate of \( 1 + i \) is \( 1 - i \). Conjugates are instrumental in simplifying complex fractions. When you multiply a complex number by its conjugate, the imaginary parts cancel out to give a real number. This technique is essential for dividing complex numbers because it eliminates the imaginary unit from the denominator, resulting in a simpler form \( a + bi \). So, anytime you see a quotient of complex numbers, think about using the conjugate for simplification.
Imaginary Unit
The imaginary unit, denoted as \( i \), is a fundamental building block of complex numbers. It is defined as the square root of \(-1\), which means \( i^2 = -1 \). This unique property allows imaginary numbers to stretch the number system beyond the real numbers. When working with complex numbers, keep in mind this characteristic of \( i \) because it crucially influences calculations, like when expanding expressions or simplifying products. Just remember, if ever you square \( i \), you get \(-1\), and this plays a key role in many aspects of complex number manipulation.
Difference of Squares
The concept of difference of squares is a powerful algebraic tool that simplifies expressions involving squares. In the case of complex numbers, the product \((1+i)(1-i)\) results in a difference of squares. This is because it follows the form \( a^2 - b^2 \), where \( a = 1 \) and \( b = i \). When you multiply, you get:
  • \( 1^2 = 1 \)
  • \( (i)^2 = -1 \)
Thus, \( 1^2 - (i)^2 = 1 + 1 = 2 \). The imaginary components cancel out, simplifying the denominator to 2, making calculations much more straightforward.
Multiplying Complex Numbers
Multiplying complex numbers follows the distributive property, just like multiplication of polynomials. Consider the expression \((3 + 5i)(1 - i)\). Start by expanding:
  • \(3 \cdot 1 = 3\)
  • \(3 \cdot (-i) = -3i\)
  • \(5i \cdot 1 = 5i\)
  • \(5i \cdot (-i) = -5i^2\)
Since \( i^2 = -1 \), replace \(-5i^2\) with \(5\). Combine all terms: \(3 - 3i + 5i + 5 = 8 + 2i\). This gives the numerator as \( 8 + 2i \). Be meticulous with each step, and remember that rearranging terms by real and imaginary components will help in writing the final expression in the standard form \( a + bi \).

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