/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Evaluate the quotient, and write... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the quotient, and write the result in the form \(a+b i\) $$\frac{2-3 i}{1-2 i}$$

Short Answer

Expert verified
The quotient is \( \frac{8}{5} + \frac{1}{5}i \).

Step by step solution

01

Identify the problem

The problem requires evaluating the complex quotient \( \frac{2-3i}{1-2i} \) and expressing the result in the form \( a + bi \).
02

Multiply by the conjugate

To eliminate the imaginary part from the denominator, multiply both the numerator and the denominator by the conjugate of the denominator, \( 1 + 2i \).\[\frac{2-3i}{1-2i} \times \frac{1+2i}{1+2i} \]
03

Simplify the denominator

Use the formula \( (a - bi)(a + bi) = a^2 + b^2 \) to simplify the denominator. Here: \[(1 - 2i)(1 + 2i) = 1^2 - (2i)^2 = 1 + 4 = 5\]
04

Expand the numerator

Distribute the terms in the numerator: \[(2-3i)(1+2i) = 2(1) + 2(2i) + (-3i)(1) + (-3i)(2i) = 2 + 4i - 3i - 6i^2\]
05

Simplify the numerator

Recall that \( i^2 = -1 \). Substitute and combine like terms in the numerator:\[2 + 4i - 3i + 6 = 8 + i\]
06

Divide to find final result

Divide the simplified numerator by the denominator to find the result in the form \( a + bi \):\[\frac{8+i}{5} = \frac{8}{5} + \frac{1}{5}i\]
07

Write the answer

The final result, written in the form \( a + bi \), is \( \frac{8}{5} + \frac{1}{5}i \).

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Key Concepts

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

Quotient of Complex Numbers
When dividing complex numbers, such as finding the quotient of \( \frac{2-3i}{1-2i} \), the goal is to express it in the form \( a + bi \). We need to make the denominator a real number to simplify the division process. This is achieved by multiplying both the numerator and the denominator by the complex conjugate of the denominator. The conjugate method ensures we eliminate the imaginary part from the denominator.

This technique leverages the fact that multiplying a complex number by its conjugate results in a real number. So, when you have a complex division task, always remember to find the conjugate of the denominator. Multiply both the numerator and the denominator by it, and simplify to get your result.
Complex Conjugate
The concept of the complex conjugate is central to dealing with complex numbers in division. A complex number is typically expressed in the form \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part. The conjugate of this complex number is \( a - bi \).

The beauty of the conjugate lies in its ability to transform complex division into a simpler calculation. For the denominator \( 1 - 2i \), the conjugate is \( 1 + 2i \). Multiplying a complex number by its conjugate results in a real number, which is the same as the sum of the squares of its components:
  • \( a^2 + b^2 \)
This is because the imaginary parts cancel each other out. Therefore, the use of the conjugate simplifies the problem, leading directly to an expression suitable for division.
Imaginary Unit
The imaginary unit, denoted as \( i \), is defined by the property \( i^2 = -1 \). It forms the foundation for all imaginary numbers, which are numbers of the form \( bi \), where \( b \) is a real number. In complex arithmetic, the imaginary unit is crucial because it allows for the extension of real numbers into two dimensions, thereby enabling the exploration of more complex mathematical concepts.

When simplifying expressions involving complex numbers, it is important to remember that any occurrence of \( i^2 \) must be replaced by \(-1\). This property is used in our provided solution during the simplification of the numerator from \( 2 + 4i - 3i - 6i^2 \) to \( 8 + i \), where \(-6i^2\) simplifies to \(6\) because \( i^2 = -1 \). This straightforward yet essential substitution helps untangle expressions and makes complex number calculations manageable.

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Most popular questions from this chapter

Gravity If an imaginary line segment is drawn between the centers of the earth and the moon, then the net gravitational force \(F\) acting on an object situated on this line segment is $$F=\frac{-K}{x^{2}}+\frac{0.012 K}{(239-x)^{2}}$$ where \(K>0\) is a constant and \(x\) is the distance of the object from the center of the earth, measured in thousands of miles. How far from the center of the earth is the "dead spot" where no net gravitational force acts upon the object? (Express your answer to the nearest thousand miles.) PICTURE CANT COPY

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DISCOVER - PROVE: Relationship Between Solutions and Coefficients The Quadratic Formula gives us the solutions of a quadratic equation from its coefficients. We can also obtain the coefficients from the solutions. (a) Find the solutions of the equation \(x^{2}-9 x+20=0\) and show that the product of the solutions is the constant term 20 and the sum of the solutions is \(9,\) the negative of the coefficient of \(x\) (b) Show that the same relationship between solutions and coefficients holds for the following equations:$$ \begin{array}{l}x^{2}-2 x-8=0 \\\x^{2}+4 x+2=0\end{array}$$ (c) Use the Quadratic Formula to prove that in general, if the equation \(x^{2}+b x+c=0\) has solutions \(r_{1}\) and \(r_{2}\) then \(c=r_{1} r_{2}\) and \(b=-\left(r_{1}+r_{2}\right)\)

Solve the equation for the variable \(x\). The constants \(a\) and \(b\) represent positive real numbers. $$a^{3} x^{3}+b^{3}=0$$

Simplify the expression. (a) \(\left(x^{-5} y^{1 / 3}\right)^{-3 / 5}\) (b) \(\left(4 r^{8} s^{-1 / 2}\right)^{1 / 2}\left(32 s^{-5 / 4}\right)^{-1 / 5}\)

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