Problem 47 Show that multiplication of comp... [FREE SOLUTION]

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Precalculus A Prelude to Calculus

Sheldon Axler

$Math Studyset 91影视 Explanations$ Math

2 Edition

Chapter 6: Problem 47

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

Short Answer

Expert verified

To show that multiplication of complex numbers is associative, we represent $u, w,$ and $z$ as $a+bi, c+di,$ and $e+fi$, respectively. Then, we calculate $(uw)z$ and $u(wz)$ as $[(ac-bd)e-(ad+bc)f] + [(ac-bd)f+(ad+bc)e]i$ and $[a(ce-df)-b(cf+de)] + [a(cf+de)+b(ce-df)]i$, respectively. After simplifying, both expressions are equal, proving that $u(wz)=(uw)z$, and thus, multiplication of complex numbers is associative.

Step by step solution

Represent the complex numbers

Let's represent the complex numbers $u, w$, and $z$ as follows: \[u=a+bi,\] \[w=c+di,\] \[z=e+fi,\] where $a, b, c, d, e, f$ are real numbers and $i$ is the imaginary unit.

Multiply $(u w) z$

First, we need to find the product of $u$ and $w$. Using the distributive property, we get: \[u w=(a+bi)(c+di)=(ac-bd)+(ad+bc)i.\] Now, let's multiply the product $(uw) = (ac-bd)+(ad+bc)i$ with $z= e+fi$. Using the distributive property again, we get: \[(u w) z = \left[(ac-bd+(ad+bc)i\right)(e+fi).\] Expanding, we obtain: \[(u w) z = [(ac-bd)e-(ad+bc)f] + [(ac-bd)f+(ad+bc)e]i.\]

Multiply $u(w z)$

Now, let's find the product of $w$ and $z$: \[w z = (c+di)(e+fi) = (ce-df) + (cf+de)i.\] Now, let's multiply the product $(wz) = (ce-df)+(cf+de)i$ with $u= a+bi$. Using the distributive property again, we get: \[u(w z) = \left[a+bi\right]\left[(ce-df)+(cf+de)i\right).\] Expanding, we obtain: \[u(w z) = [a(ce-df)-b(cf+de)] + [a(cf+de)+b(ce-df)]i.\]

Compare the results

Now we compare the results of the multiplications in Steps 2 and 3. We notice that, after simplifying, both expressions are equal: \[(u w) z = [(ac-bd)e-(ad+bc)f] + [(ac-bd)f+(ad+bc)e]i,\] \[u(w z) = [a(ce-df)-b(cf+de)] + [a(cf+de)+b(ce-df)]i.\] After rearranging the terms, we can observe that: \[((ac-bd)e-(ad+bc)f) = (a(ce-df)-b(cf+de)),\] and: \[((ac-bd)f+(ad+bc)e) = (a(cf+de)+b(ce-df)).\] Hence, we can conclude that multiplication of complex numbers is associative, as $(uw)z = u(wz)$ for all complex numbers $u, w$, and $z$.

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91影视

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

Short Answer

Step by step solution

Represent the complex numbers

Multiply \((u w) z\)

Multiply \(u(w z)\)

Compare the results

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