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Chapter 6: Problem 46

Show that multiplication of complex numbers is commutative, meaning that $$ w z=z w $$ for all complex numbers \(w\) and \(z\).

Short Answer

Expert verified
To show that the multiplication of complex numbers is commutative, let \(w = a + bi\) and \(z = c + di\), where \(a, b, c, d\) are real numbers. Compute the product in standard (wz) and reverse (zw) order: \(wz = (ac - bd) + (ad + bc)i\) \(zw = (ca - db) + (cb + da)i\) Since \(ac - bd = ca - db\) and \(ad + bc = cb + da\), we find that \(wz = zw\). Thus, multiplication of complex numbers is commutative.

Step by step solution

01

Define Complex Numbers

Let \(w = a + bi\) and \(z = c + di\), where \(a, b, c, d\) are real numbers, and \(i\) is the imaginary unit with the property \(i^2 = -1\).
02

Multiply Complex Numbers in the Standard Order (wz)

We start by multiplying \(w\) and \(z\) in the standard order: \(wz = (a + bi)(c + di) = ac + adi + bci + bdi^2\) Notice that since \(i^2 = -1\), we can replace \(i^2\) with \(-1\): \(wz = ac + adi + bci - bd\) Now, we will rearrange the terms: \(wz = (ac - bd) + (ad + bc)i\)
03

Multiply Complex Numbers in the Reverse Order (zw)

Now, let's multiply \(z\) and \(w\) in reverse order and compare the result to the previous calculation: \(zw = (c + di)(a + bi) = ca + cbi + dai + dbi^2\) As before, we can replace \(i^2\) with \(-1\): \(zw = ca + cbi + dai - db\) Now, we will rearrange the terms: \(zw = (ca - db) + (cb + da)i\)
04

Compare the Results

We can now compare the results of the two multiplications. Notice that: \(wz = (ac - bd) + (ad + bc)i\) \(zw = (ca - db) + (cb + da)i\) Since \(ac - bd = ca - db\) and \(ad + bc = cb + da\), we can conclude that: \(wz = zw\) Therefore, multiplication of complex numbers is commutative.

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Most popular questions from this chapter

Describe the subset of the complex plane consisting of the complex numbers \(z\) such that the real part of \(z^{3}\) is a positive number.

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