Problem 60
Question
Write the quotient in standard form. $$\frac{8-7 i}{1-2 i}$$
Step-by-Step Solution
Verified Answer
The standard form of the quotient is \(2.8+1.8i\).
1Step 1: Identify the numerator and the denominator
The numerator is \(8-7i\) and the denominator is \(1-2i\).
2Step 2: Calculate the conjugate of the denominator
The conjugate of \(1-2i\) is \(1+2i\).
3Step 3: Multiply the numerator and the denominator by the conjugate of the denominator
The equation becomes \(\frac{(8-7i)(1+2i)}{(1-2i)(1+2i)}\).
4Step 4: Distribute and multiply
Carry out the multiplication in the numerator and the denominator. In the numerator, the result is \(8+16i-7i-14i^2\). In the denominator, we get a difference of squares: \((1)^2-(2i)^2\).
5Step 5: Simplify
Simplify the numerator and the denominator. \(i^2\) is \(-1\), so \(-14i^2\) becomes \(14\). So, the numerator is \(14+9i\), and the denominator is \(1-(-4)\), which simplifies to 5. Therefore, the complex number in standard form is \(\frac{14+9i}{5}\).
6Step 6: Divide the real and imaginary parts by the denominator
Divide the real part 14 by 5 and the imaginary part 9 by 5 to get \(2.8+1.8i\).
Key Concepts
Conjugate of Complex NumbersDistributive Property in AlgebraSimplifying Complex Expressions
Conjugate of Complex Numbers
The 'conjugate' of a complex number is an important concept to understand when working with complex arithmetic. Given a complex number in the form of \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) represents the imaginary unit, the conjugate of this number is \(a - bi\). Essentially, you just change the sign of the imaginary part.
For example, if we take the complex number \(1 - 2i\), its conjugate is \(1 + 2i\). This operation is useful because multiplying a complex number by its conjugate gives a real number, a trick often used to simplify the division of complex numbers as seen in the textbook exercise. Multiplying both the numerator and the denominator of a complex fraction by the conjugate of the denominator removes the imaginary part from the denominator, making further simplification possible.
For example, if we take the complex number \(1 - 2i\), its conjugate is \(1 + 2i\). This operation is useful because multiplying a complex number by its conjugate gives a real number, a trick often used to simplify the division of complex numbers as seen in the textbook exercise. Multiplying both the numerator and the denominator of a complex fraction by the conjugate of the denominator removes the imaginary part from the denominator, making further simplification possible.
Distributive Property in Algebra
The distributive property is a fundamental principle in algebra that allows us to multiply a single term by each term within a parenthesis in an expression. It is expressed mathematically as \(a(b + c) = ab + ac\). This property makes it possible to expand expressions and is utilized to perform operations like multiplication of complex numbers.
In the context of complex numbers, when you distribute to multiply, you must consider the imaginary unit \(i\) as well, which has the property that \(i^2 = -1\). This is why, during multiplication, when we encounter terms like \(-7i \times 2i\), we end up with \(-14i^2\), which simplifies to \(+14\) because of this unique property of the imaginary unit.
In the context of complex numbers, when you distribute to multiply, you must consider the imaginary unit \(i\) as well, which has the property that \(i^2 = -1\). This is why, during multiplication, when we encounter terms like \(-7i \times 2i\), we end up with \(-14i^2\), which simplifies to \(+14\) because of this unique property of the imaginary unit.
Simplifying Complex Expressions
When simplifying complex expressions, especially after using the distributive property to expand them, we combine like terms and simplify using \(i^2 = -1\). In our example, \(8+16i-7i-14i^2\) becomes \(8 - 14 + (16i - 7i)\), simplifying to \(14 + 9i\) after applying the property of the imaginary unit to the \(-14i^2\) term.
Solutions should present complex numbers in standard form, which is \(a + bi\). If we have a quotient, like in this exercise, we divide both the real and the imaginary parts by the denominator. It's important, as shown in Step 6 of the solution, to individually divide and simplify the real part (14 divided by 5 equals 2.8) and the imaginary part (9 divided by 5 equals 1.8), resulting in the standard form of \(2.8 + 1.8i\).
Solutions should present complex numbers in standard form, which is \(a + bi\). If we have a quotient, like in this exercise, we divide both the real and the imaginary parts by the denominator. It's important, as shown in Step 6 of the solution, to individually divide and simplify the real part (14 divided by 5 equals 2.8) and the imaginary part (9 divided by 5 equals 1.8), resulting in the standard form of \(2.8 + 1.8i\).
Other exercises in this chapter
Problem 60
Use the Quadratic Formula to solve the equation. Use a graphing utility to verify your solutions graphically. $$x^{2}+16=-5 x$$
View solution Problem 60
Use a graphing utility to approximate any solutions of the equation. [Remember to write the equation in the form \(f(x)=0.1\) $$8-\sqrt{x+9}=6$$
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Solving an Equation Involving an Absolute Value Find all solutions of the equation algebraically. Check your solutions. $$|x|=x^{2}+x-24$$
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Determine the intervals on which the polynomial is entirely negative and those on which it is entirely positive. $$-x^{2}+6 x-10$$
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