/*! 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 76 Evaluate the given expression fo... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the given expression for \(z=3-4 i\) and \(w=5+2 i\) $$\bar{z} \cdot \bar{w}$$

Short Answer

Expert verified
\(\bar{z} \cdot \bar{w} = 23 + 14i\).

Step by step solution

01

Understand the problem

We need to evaluate the expression \(\bar{z} \cdot \bar{w}\) given that \(z = 3 - 4i\) and \(w = 5 + 2i\). Here, \(\bar{z}\) and \(\bar{w}\) denote the complex conjugates of \(z\) and \(w\) respectively. The complex conjugate of a complex number \(a + bi\) is \(a - bi\).
02

Find the complex conjugates

Calculate the complex conjugates of \(z\) and \(w\). - The conjugate of \(z = 3 - 4i\) is \(\bar{z} = 3 + 4i\).- The conjugate of \(w = 5 + 2i\) is \(\bar{w} = 5 - 2i\).
03

Multiply the complex conjugates

To find \(\bar{z} \cdot \bar{w}\), multiply \(\bar{z} = 3 + 4i\) and \(\bar{w} = 5 - 2i\):\[(3 + 4i)(5 - 2i) = 3 \times 5 + 3 \times (-2i) + 4i \times 5 + 4i \times (-2i)\]Calculating each term, we have:- \(3 \times 5 = 15\)- \(3 \times (-2i) = -6i\)- \(4i \times 5 = 20i\)- \(4i \times (-2i) = -8i^2 = 8\) since \(i^2 = -1\).
04

Simplify the expression

Combine the terms from the multiplication:\[15 - 6i + 20i + 8 = (15 + 8) + (-6i + 20i) = 23 + 14i\].

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Key Concepts

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

Complex Conjugate
A complex conjugate is a companion of a complex number with a subtle alteration. For any given complex number of the form \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit, the complex conjugate is \(a - bi\).

This transformation simply flips the sign in front of the imaginary part. The role of the complex conjugate is crucial in various operations, especially in division and multiplication of complex numbers, where conjugates help to simplify expressions and get rid of imaginary components in denominators.

It's like substituting one half of a balanced equation, but instead of changing a part of the equation, you're flipping the sign of the imaginary portion. This process often serves to make calculations with complex numbers more manageable, like when finding the product of two conjugates results in a real number.
Multiplication of Complex Numbers
Multiplying complex numbers might seem daunting, but it follows a straightforward rule akin to multiplying polynomials. If you have two complex numbers, \(z = a + bi\) and \(w = c + di\), you multiply them just like you would multiply two binomials using the distributive property (also known as the FOIL method):
  • First multiply \(a\) and \(c\)
  • Outside with \(a\) and \(di\)
  • Inside \(bi\) and \(c\)
  • Last \(bi\) and \(di\)

The important thing to remember when doing these operations is to apply the property \(i^2 = -1\), which changes the imaginary unit into a negative one.

This step is key to eliminating the imaginary unit from certain parts of the resulting term, producing a combination of real and imaginary numbers like in the product example, \((3 + 4i)(5 - 2i) = 23 + 14i\). Notice how the terms are combined:
  • Real parts add up after multiplication
  • Imaginary parts are also added together

Multiplying complex conjugates, like with our example of \(\bar{z}\) and \(\bar{w}\), simplifies calculations as the imaginary parts can cancel each other out, leaving a purely real number, helpful in complex arithmetic.
Imaginary Unit
The imaginary unit \(i\) is an interesting and essential component of complex numbers. It is defined as \(i = \sqrt{-1}\), meaning \(i^2 = -1\). This intriguing property forms the foundation of working with imaginary numbers and complex arithmetic.

Imaginary numbers arise from dealing with square roots of negative numbers, which aren't possible within the set of real numbers. By introducing \(i\), we can extend real numbers and perform operations with these complex numbers efficiently.

In practical terms, knowing how to handle \(i\) ensures you can perform operations like multiplication and division in the complex plane without hassle.

Since \(i^2 = -1\), when you encounter \(i^2\), you can simplify it by replacing it with \(-1\). This approach helps negate certain terms or transform imaginary components into real numbers, as seen when multiplying complex numbers. It's like a quick conversion trick within computations to ensure outputs remain balanced between their real and imaginary parts.

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Most popular questions from this chapter

Fish Population The fish population in a certain lake rises and falls according to the formula $$F=1000\left(30+17 t-t^{2}\right)$$ Here \(F\) is the number of fish at time \(t,\) where \(t\) is measured in years since January \(1,2002,\) when the fish population was first estimated. (a) On what date will the fish population again be the same as it was on January \(1,2002 ?\) (b) By what date will all the fish in the lake have died?

Radicals Simplify the expression, and eliminate any negative exponents(s). Assume that all letters denote positive numbers. (a) \(\sqrt[3]{y \sqrt{y}}\) (b) \(\sqrt{\frac{16 u^{3} v}{u v^{5}}}\)

Radicals Simplify the expression, and eliminate any negative exponents(s). Assume that all letters denote positive numbers. (a) \(\sqrt{s \sqrt{s^{3}}}\) (b) \(\sqrt[3]{\frac{54 x^{2} y^{4}}{2 x^{5} y}}\)

Use scientific notation, the Laws of Exponents, and a calculator to perform the indicated operations. State your answer rounded to the number of significant digits indicated by the given data. $$\left(1.062 \times 10^{24}\right)\left(8.61 \times 10^{19}\right)$$

It follows from Kepler's Third Law of planetary motion that the average distance from a planet to the sun (in meters) is $$d=\left(\frac{G M}{4 \pi^{2}}\right)^{1 / 3} T^{2 / 3}$$ where \(M=1.99 \times 10^{30} \mathrm{kg}\) is the mass of the sun, \(G=6.67 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\) is the gravitational constant, and \(T\) is the period of the planet's orbit (in seconds). Use the fact that the period of the earth's orbit is about 365.25 days to find the distance from the earth to the sun.

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