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Chapter 12: Problem 36

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) \((1+i)^{2}+(1+i)\) (b) \((1+3 i)^{2}+(1+i)\) (c) \((-7+7 i)^{2}+(1+i)\)

Short Answer

Expert verified
The simplified forms of the expressions are (a) \( 1+3i \), (b) \( -7+7i \), and (c) \( 1-97i \).

Step by step solution

01

Break down the first expression

We are asked to simplify the expression \( (1+i)^{2}+(1+i)\). We first square the complex number (1+i), which means multiplying it by itself: \[ (1+i)(1+i) = 1 + i + i + i^{2} = 1 + 2i -1 \] since \( i^{2} = -1 \). We then simplify it to obtain \( 2i \). We then add this to the remaining part of the expression along with real part and imaginary part separately: \( 2i + (1+i) = 1 + 3i\)
02

Break down the second expression

The next expression to simplify is \((1+3i)^{2} + (1+i)\). We follow the same steps as in Step 1. First, we square \( (1+3i) \): \[ (1+3i)(1+3i) = 1 + 3i + 3i + 9i^{2} = 1 + 6i - 9 \] because \( i^{2} = -1 \), which simplifies to \( -8 + 6i \). We then add \( (1+i) \) to this: \( -8 + 6i + 1 + i = -7+7i \).
03

Break down the third expression

The final expression to simplify is \( (-7+7i)^{2} + (1+i) \). First, we square \( -7+7i \): \[ (-7+7i)(-7+7i) = 49 - 49i - 49i + 49i^{2} = 49 - 98i -49 \] because \( i^{2} = -1 \), which simplifies to \( 0 - 98i \). We then add \( (1+i) \) to this: \( - 98i + 1 + i = 1 - 97i \).

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Key Concepts

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

Arithmetic Operations with Complex Numbers
Working with complex numbers involves treating them as expressions composed of real and imaginary parts. A complex number looks like \(a + bi\), where \(a\) is the real part and \(bi\) is the imaginary part.

When adding two complex numbers, simply add their corresponding real parts and imaginary parts.
  • For example, to add \((2+3i)\) and \((5+2i)\), you combine real parts \(2 + 5 = 7\) and imaginary parts \(3i + 2i = 5i\).
Multiplication feels a bit like expanding polynomials, using the rule \(i^2 = -1\).
  • When multiplying \((2+3i)(5+2i)\), you calculate it as \(10 + 4i + 15i + 6i^2\). After replacing \(i^2\) with \(-1\), it simplifies to \(4 + 19i\).
This understanding helps you handle complex expressions with confidence.
Squaring Complex Numbers
Squaring complex numbers is an interesting operation. It's like multiplying the complex number by itself.

Consider the complex number \((a+bi)\). When squaring it, you perform \((a+bi)(a+bi)\).
  • This expansion leads to \(a^2 + 2abi + b^2i^2\).
  • Since \(i^2 = -1\), replace \(b^2i^2\) with \(-b^2\).
  • The result is \(a^2 - b^2 + 2abi\).
In our examples:
  • \((1+i)^2 = 1 + 2i - 1 = 2i\).
  • \((1+3i)^2 = 1 + 6i - 9 = -8 + 6i\).
  • \((-7+7i)^2 = 49 - 98i - 49 = -98i\).
Recognizing how to square complex numbers helps simplify expressions.
Imaginary Unit
The imaginary unit \(i\) is a key concept in complex numbers. It's defined such that \(i^2 = -1\).

This might seem strange at first, as it doesn't have a real number equivalent, but it allows us to extend the number system into the complex plane.
  • \(i\) is used to represent imaginary parts of complex numbers like \(a + bi\), where \(b\) is a real number.
  • When working with complex equations, knowing that \(i^2 = -1\) simplifies multiplication and squaring.
For example:
  • In \((2+3i)(5+2i)\), \(i^2\) helps reduce expressions, effectively turning them into real numbers.
  • During the squaring of complex numbers like \((1+i)^2\), \(i^2\) allows us to reframe \(i^2\) as \(-1\).
Understanding this imaginary unit is crucial for computations and simplifications in complex numbers.

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Most popular questions from this chapter

Consider the Mandelbrot sequence with seed \(s=-0.25\). (a) Using a calculator find \(s_{1}\) through \(s_{10}\), rounded to six decimal places. (b) Suppose you are given \(s_{N}=-0.207107\). Using a calculator find \(s_{N+1}\), rounded to six decimal places. (c) Is this Mandelbrot sequence escaping, periodic, or attracted? Explain.

Refer to the Sierpinski ternary gasket, \(a\) variation of the Sierpinski gasket defined by the following recursive replacement rule. Assume that the seed triangle of the Sierpinski ternary gasket has area \(A=81 .\) 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 \(12-18\). $$ \begin{array}{l|l|l|l|l} & R & S & T & Q \\ \hline \text { Start } & 0 & 0 & 0 & 81 \\ \hline \text { Step 1 } & 3 & 9 & 27 & 54 \\ \hline \text { Step 2 } & & & & \\ \hline \text { Step 3 } & & & & \\ \hline \text { Step 4 } & & & & \\ \hline \text { Step } \boldsymbol{N} & & & & \end{array} $$

Consider the Mandelbrot sequence with seed \(s=-1.25 .\) Is this Mandelbrot sequence escaping, periodic, or attracted? If attracted, to what number?

Consider the Mandelbrot sequence with seed \(s=-i\). (a) Find \(s_{1}\) through \(s_{5}\). (Hint: Try Exercise 35 first.) (b) Is this Mandelbrot sequence escaping, periodic, or attracted? Explain.

Show that the Mandelbrot set has a reflection symmetry. (Hint: Compare the Mandelbrot sequences with seeds \(a+b i\) and \(a-b i .)\)
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