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Chapter 12: Problem 35

Are a review of complex number arithmetic. Recall that (1) to add two complex numbers you simply add the real parts and the imaginary parts: e.g., \((2+3 i)+(5+2 i)=\) \(7+5 i ;\) (2) to multiply two complex numbers you multiply them as if they were polynomials and use the fact that \(i^{2}=-1:\) e.g., \((2+3 i)(5+2 i)=10+4 i+15 i+6 i^{2}=4+19 i .\) Finally, if you know how to multiply two complex numbers then you also know how to square them, since \((a+b i)^{2}=(a+b i)(a+b i)\). Simplify each expression. (a) \((-i)^{2}+(-i)\) (b) \((-1-i)^{2}+(-i)\) (c) \(i^{2}+(-i)\)

Short Answer

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The answers are (a) \(-1 - i\), (b) \(i\), and (c) \(-1 - i\).

Step by step solution

01

Simplifying the first expression

Given is the expression \((-i)^{2}+(-i)\). Square the \(-i\) first: \((-i)^{2} = -1\). Then, add \(-i\) to it. The resulting expression is \(-1 - i\).
02

Simplifying the second expression

For the expression \((-1-i)^{2}+(-i)\), we start by squaring \(-1-i\). Utilizing the rule \((a+b i)^{2}=(a+b i)(a+b i)\), we get \(1 + 2i + i^{2}\). Since \(i^{2} = -1\), simplify it to \(1 + 2i - 1\). Now, adding \(-i\) to this expression, we get \(2i - i = i\).
03

Simplifying the third expression

Lastly, for the expression \(i^{2}+(-i)\), replace \(i^{2}\) by \(-1\) (given that \(i^{2} = -1\)). Upon simplification, we get \(-1 - i\).

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Key Concepts

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

Complex Number Arithmetic
In mathematics, complex number arithmetic involves performing operations like addition, subtraction, multiplication, and division with complex numbers. A complex number looks like this: \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit, representing \(\sqrt{-1}\). Here's a quick breakdown of how to perform these operations:
  • Addition: Add the real parts together and the imaginary parts together. For example, \((2 + 3i) + (5 + 2i) = 7 + 5i\).
  • Subtraction: Subtract the real and imaginary parts respectively. For instance, \((5 + 6i) - (2 + 3i) = 3 + 3i\).
  • Multiplication: Multiply the numbers as if they are two binomials, then use \(i^2 = -1\) to simplify. For instance, \((2 + 3i)(5 + 2i) = 10 + 4i + 15i + 6(-1) = 4 + 19i\).
  • Division: Multiply the numerator and denominator with the conjugate of the denominator to eliminate the imaginary unit from the denominator.
Understanding these operations allows you to tackle problems involving complex numbers with confidence.
Remember, complex numbers are extensions of the real numbers and allow for solutions to equations that have no real solutions.
Imaginary Numbers
Imaginary numbers extend the number system and have their roots in the need to solve equations that have no solutions within the real numbers. The imaginary unit \(i\) is defined as the square root of \(-1\). Imaginary numbers are often represented as \(bi\), where \(b\) is a real number. Here are some key facts about imaginary numbers:
  • Imaginary Unit: The imaginary unit \(i\) satisfies \(i^2 = -1\). This is the fundamental property used in many computations involving complex numbers.
  • Powers of \(i\): The powers of \(i\) cycle through four stages: \(i, -1, -i,\) and \(1\). For example, \(i^3 = -i\) and \(i^4 = 1\).
  • Applications: Imaginary numbers are used in various fields such as engineering, physics, and signal processing to represent phenomena that are not easily described using only real numbers.
Recognizing the role of imaginary numbers allows you to solve otherwise unsolvable equations and to work alongside complex numbers with ease.
Polynomial Multiplication
Polynomial multiplication, in the context of complex numbers, involves multiplying complex expressions similar to how you would multiply polynomials. Consider complex numbers as binomials where the operations follow basic algebraic rules:
  • Distributive Property: You distribute each term in the first polynomial by each term in the second, similar to the FOIL (First, Outer, Inner, Last) method used for binomials.
  • Simplification: After expansion, look for terms involving \(i^2\) since \(i^2 = -1\) to simplify the resulting expression.
  • Strategy: By viewing complex number multiplication as polynomial multiplication, it becomes easier to handle complex expressions such as \((a + bi)(c + di)\).
Working through problems with polynomial multiplication builds a strong foundation in manipulating complex numbers and aids in understanding their underlying structure.

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Most popular questions from this chapter

Refer to \(a\) variation of the Koch snowflake called the quadratic Koch fractal. The construction of the quadratic Koch fractal is similar to that of the Koch snowflake, but it uses squares instead of equilateral triangles as the shape's building blocks. The following recursive construction rule defines the quadratic Koch fractal: Start. Start with a solid seed square [Fig. \(12-35(a)]\). Step 1. Attach a smaller square (sides one-third the length of the sides of the seed square) to the middle third of each side [Fig. \(12-35(b)]\). Step 2. Attach a smaller square (sides one-third the length of the sides of the previous side to the middle third of each side [Fig. \(12-35(\mathrm{c})] .\) (Call this procedure \(Q K F .)\) Steps \(3,4,\) etc. At each step, apply procedure \(Q K F\) to the figure obtained in the preceding step. Assume that the seed square of the quadratic Koch fractal has area \(A=243 .\) Let \(R\) denote the number of squares added at a particular step, \(S\) the area of each added square, \(T\) the total new area added, and \(Q\) the area of the shape obtained at a particular step of the construction. Complete the missing entries in Table \(12-8\). $$ \begin{array}{l|c|c|c|c} & R & S & T & Q \\ \hline \text { Start } & 0 & 0 & 0 & 243 \\ \hline \text { Step 1 } & 4 & 27 & 108 & 351 \\ \hline \text { Step 2 } & 20 & 3 & 60 & 411 \\ \hline \text { Step 3 } & & & & \\ \hline \text { Step 4 } & & & & \\ \hline \end{array} $$

Refer to a variation of the Koch snowflake called the Koch antisnowflake. The Koch antisnowflake is much like the Koch snowflake, but it is based on a recursive rule that removes equilateral triangles. The recursive replacement rule for the Koch antisnowflake is as follows: Assume that the seed triangle of the Koch antisnowflake has sides of length \(18 \mathrm{~cm} .\) Let \(M\) denote the number of sides, \(L\) the length of each side, and \(P\) the perimeter of the shape obtained at the indicated step of the construction. Complete the missing entries in Table \(12-10 .\) $$ \begin{array}{l|c|c|c} & M & L & P \\ \hline \text { Start } & 3 & 18 \mathrm{~cm} & 54 \mathrm{~cm} \\ \hline \text { Step 1 } & 12 & 6 \mathrm{~cm} & 72 \mathrm{~cm} \\ \hline \text { Step 2 } & & & \\ \hline \text { Step 3 } & & & \\ \hline \text { Step 4 } & & & \\ \hline \text { Step 5 } & & & \\ \hline \end{array} $$

Refer to the Menger sponge, \(a\) three dimensional cousin of the Sierpinski gasket. The Menger sponge is defined by the following recursive construction rule. Start. Start with a solid seed cube [Fig. \(12-42(a)\) ]. Step 1. Subdivide the seed cube into 27 equal subcubes and remove the central cube and the six cubes in the centers of each face. This leaves \(a\) "sponge" consisting of 20 solid subcubes, as shown in Fig. \(12-42(b) .\) Step 2. Subdivide each solid subcube into 27 subcubes and remove the central cube and the six cubes in the centers of each face. This gives the "sponge" shown in Fig. \(12-42(\mathrm{c}) .\) (Call the procedure of removing the central cube and the cubes in the center of each face procedure MS.) Steps \(3,4,\) etc. Apply procedure MS to each cube of the "sponge" obtained in the previous step. Assume that the seed cube of the Menger sponge has volume 1 . (a) Let \(C\) denote the total number of cubes removed at a particular step of the construction, \(U\) the volume of each removed cube, and \(V\) the volume of the sponge at that particular step of the construction. Complete the entries in the following table. $$ \begin{array}{l|l|l|l} & C & U & V \\ \hline \text { Start } & 0 & 0 & 1 \\ \hline \text { Step 1 } & 7 & \frac{1}{27} & \frac{20}{27} \\ \hline \text { Step 2 } & & & \\ \hline \text { Step 3 } & & & \\ \hline \text { Step 4 } & & & \\ \hline \text { Step } \boldsymbol{N} & & & \end{array} $$

Refer to the construction of the quadratic Koch island. The quadratic Koch island is defined by the following recursive replacement rule. Start: Start with a seed square [Fig. \(12-37(a)]\). (Notice that here we are only dealing with the boundary of the square.) Replacement rule: In each step replace any horizontal boundary segment with the "sawtooth" version shown in Fig. \(12-37(b)\) and any vertical line segment with the "sawtooth" version shown in Fig. \(12-37(c)\). Consider the construction of a Sierpinski gasket starting with a seed triangle of area \(A=1\). Let \(R\) denote the number of triangles removed at a particular step, \(S\) the area of each removed triangle, \(T\) the total area removed, and \(Q\) the area of the "gasket" obtained at a particular step of the construction. Complete the missing entries in Table \(12-14\) $$ \begin{array}{l|l|l|l|l} & R & S & T & Q \\ \hline \text { Start } & 0 & 0 & 0 & 1 \\ \hline \text { Step 1 } & 1 & \frac{1}{4} & \frac{1}{4} & \frac{3}{4} \\ \hline \text { Step 2 } & 3 & \frac{1}{16} & \frac{3}{16} & \frac{9}{16} \\ \hline \text { Step 3 } & & & & \\ \hline \text { Step 4 } & & & & \\ \hline \text { Step 5 } & & & & \end{array} $$

Consider the construction of a Koch snowflake starting with a seed triangle having sides of length \(81 \mathrm{~cm} .\) Let \(M\) denote the number of sides, \(L\) the length of each side, and \(P\) the perimeter of the "snowflake" obtained at the indicated step of the construction. Complete the missing entries in Table \(12-1\) $$ \begin{array}{l|c|c|c} & M & L & P \\ \hline \text { Start } & 3 & 81 \mathrm{~cm} & 243 \mathrm{~cm} \\ \hline \text { Step 1 } & 12 & 27 \mathrm{~cm} & 324 \mathrm{~cm} \\ \hline \text { Step 2 } & & & \\ \hline \text { Step 3 } & & & \\ \hline \text { Step 4 } & & & \\ \hline \text { Step 5 } & & & \end{array} $$
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