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Evaluate each expression using the values \(z=2+3 i, w=9-4 i,\) and \(w_{1}=-7-i\). $$(z w) w_{1}$$

Short Answer

Expert verified
\((zw)w_1 = -191 - 163i\)

Step by step solution

01

Substitute Given Values

Substitute the given values of the expressions into the formula. We have:1. \( z = 2+3i \)2. \( w = 9-4i \) 3. \( w_1 = -7-i \). The expression to evaluate is \( (zw)w_1 \).
02

Multiply z and w

Multiply the complex numbers \( z \) and \( w \): \[ (2+3i)(9-4i) \]}.
03

Apply Distributed Property

Use the distributive property to expand the multiplication:\( (2+3i)(9-4i) = 2 \cdot 9 + 2 \cdot (-4i) + 3i \cdot 9 + 3i \cdot (-4i) \).Simplify each term:- \( 2 \cdot 9 = 18 \)- \( 2 \cdot (-4i) = -8i \)- \( 3i \cdot 9 = 27i \)- \( 3i \cdot (-4i) = -12i^2 \).
04

Simplify Using \(i^2 = -1\)

Since \( i^2 = -1 \), replace \( i^2 \) with \(-1\) in the expression:\( -12i^2 = -12(-1) = 12 \).The expression becomes:\( 18 - 8i + 27i + 12 \).
05

Combine Like Terms

Combine the real and imaginary parts:Real part: \( 18 + 12 = 30 \)Imaginary part: \( -8i + 27i = 19i \)So, \( zw = 30 + 19i \).
06

Multiply the Result by w1

We now multiply the result from \( zw = 30 + 19i \) by \( w_1 = -7-i \):\[ (30 + 19i)(-7-i) \]
07

Apply Distribution Again

Use the distributive property:\( 30(-7) + 30(-i) + 19i(-7) + 19i(-i) \)Simplify each term:- \( 30 \cdot (-7) = -210 \)- \( 30 \cdot (-i) = -30i \)- \( 19i \cdot (-7) = -133i \)- \( 19i \cdot (-i) = -19i^2 \).
08

Simplify Using \(i^2 = -1\) Again

Again, using \( i^2 = -1 \), simplify:\( -19i^2 = -19(-1) = 19 \).The expression becomes:\( -210 - 30i - 133i + 19 \).
09

Combine Like Terms Again

Combine the real and imaginary parts:Real part: \( -210 + 19 = -191 \)Imaginary part: \(-30i - 133i = -163i \)Thus the evaluation gives us the final expression:\( (zw)w_1 = -191 - 163i \).

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Key Concepts

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

Distributive Property
The distributive property is a fundamental concept in algebra that allows us to expand expressions and simplify calculations. When multiplying expressions, the distributive property states that you can "distribute" the multiplication over each term inside the parentheses. In mathematical terms, if you have an expression like \((a + b)(c + d)\), the distributive property lets you rewrite it as \(a(c + d) + b(c + d)\). This is further expanded to \(ac + ad + bc + bd\).

Applying the distributive property is especially vital when dealing with complex numbers because it helps break down the multiplication into manageable parts. For instance, in the exercise, when multiplying \((2 + 3i)(9 - 4i)\), each part of the first complex number is multiplied by each part of the second complex number. By applying the distributive property, complex multiplication becomes a straightforward process of expanding and simplifying terms.

Using the property ensures accuracy and makes the transition to combination and simplification of real and imaginary parts possible. It's essential to practice using the distributive property to get comfortable with handling complex numbers and algebraic expressions.
Imaginary Unit
The imaginary unit, represented as \(i\), is a concept that extends our number system beyond the real numbers. In mathematics, \(i\) is defined as the square root of \(-1\). That might seem a bit unusual since the square root of a negative number isn't a concept we encounter in the real number system. However, introducing \(i\) allows us to work with what we call 'complex numbers.'

Complex numbers are in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit. A crucial property of \(i\) is that \(i^2 = -1\). This property is frequently used to simplify expressions during calculations, as we saw in the exercise while dealing with terms like \(-12i^2\) and \(-19i^2\). By converting these terms using \(i^2 = -1\), they transform into real numbers: \(-12i^2 = 12\) and \(-19i^2 = 19\).

Understanding the imaginary unit is key to mastering complex numbers. It allows for the expansion of algebraic concepts into a broader scope, making equations solvable that appear perplexing when restricted to the real numbers alone.
Complex Multiplication
Complex multiplication involves the multiplication of two complex numbers. This process combines both the real and imaginary components using the distributive property. As covered earlier, multiplying complex numbers is straightforward once you identify each term in your complex numbers.

For example, consider multiplying \(z = 2+3i\) and \(w = 9-4i\). To simplify this, we expand \((2+3i)(9-4i)\) using the distributive property. We multiply each part of the first number by each part of the second number, as follows:
  • \(2 \cdot 9 = 18\)
  • \(2 \cdot (-4i) = -8i\)
  • \(3i \cdot 9 = 27i\)
  • \(3i \cdot (-4i) = -12i^2\)
By collecting and simplifying these terms with the imaginary unit's property \(i^2 = -1\), we find \(zw = 30 + 19i\).

Similarly, you can multiply this result with another complex number, as in \((30 + 19i)(-7 - i)\), using the same method. Each multiplication step involves separating and recombining terms into real and imaginary parts, resulting in a final simplified product. Mastery of complex multiplication opens doors to exploring further complex analysis topics, enabling the solving of equations previously unsolvable over the reals.

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Most popular questions from this chapter

Find all roots of each equation. Hints: First, factor by grouping. In Exercises 71 and 72 each equation has three roots; in Exercise 73 the equation has six roots; in Exercise 74 there are five roots. $$x^{6}-9 x^{4}+16 x^{2}-144=0$$

Determine whether the given value is a zero of the function. \(f(x)=x^{3}-3 x^{2}+3 x-3\) (a) \(x=\sqrt[3]{2}-1\) (b) \(x=\sqrt[3]{2}+1\)

Find a quadratic equation with the given roots. Write your answers in the form \(A x^{2}+B x+C=0\) Suggestion: Make use of Table 2. $$r_{1}=1+\sqrt{5}, r_{2}=1-\sqrt{5}$$

Provides an example in which an error in a partial fraction decomposition is not easily detected with a graphical approach. Indeed, this may be an example of a case in which, to check your partial fractions work, it’s easier to repeat the algebra than to experiment with numerous viewing rectangles. Decide for yourself after completing the problem. There is an error in the following partial fraction decomposition: $$\frac{1}{(x+2)(x+5)(x-14)}=\frac{-1 / 48}{x+2}+\frac{1 / 57}{x+5}+\frac{1 / 305}{x-14}$$ (a) Let \(f\) and \(g\) denote the two functions defined by the expressions on the left side and the right side, respectively, in the above equation. Use a graphing utility to graph \(f\) and \(g,\) first in the standard viewing rectangle and then in the rectangle [-15,15,5] by \([-0.02,0.04,0.02] .\) In this latter rectangle, note that the graphs do appear to be identical. (People using a software graphing application and looking at the curves on a computer monitor may have a little advantage here over those drawing the graphs on a relatively small graphing calculator screen.) (b) Find a viewing rectangle clearly demonstrating that the graphs of \(f\) and \(g\) are not identical. (c) Find the correct partial fraction decomposition, given that the form is $$\frac{1}{(x+2)(x+5)(x-14)}=\frac{A}{x+2}+\frac{B}{x+5}+\frac{C}{x-14}$$

This exercise completes a detail mentioned in the text in the derivation of the lineariand quadratic factors theorem. Let \(a\) and \(b\) be real numbers with \(b \neq 0 .\) Show that the quadratic polynomial \(x^{2}-2 a x+a^{2}+b^{2}\) is irreducible.

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