Problem 48 Show that addition and multiplic... [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 48

Show that addition and multiplication of complex numbers satisfy the distributive property, meaning that $$ u(w+z)=u w+u z $$ for all complex numbers $u, w,$ and $z$.

Short Answer

Expert verified

The distributive property holds true for complex numbers. Given complex numbers $u = a + bi$, $w = x + yi$, and $z = c + di$, where $a, b, x, y, c, d$ are real numbers, and $i$ is the imaginary unit ($i^2 = -1$). We can show that $u(w+z) = uw + uz$ by performing the complex number operations, which results in the expressions $(ax + ac - by - bd) + (ay + ad + bx + bc)i$ for both $u(w+z)$ and $uw + uz$.

Step by step solution

Represent the complex numbers

Let $u = a + bi$, $w = x + yi$, and $z = c + di$, where $a, b, x, y, c, d$ are real numbers, and $i$ is the imaginary unit ($i^2 = -1$).

Calculate $w+z$

To find the sum of complex numbers $w$ and $z$, we simply add their respective real and imaginary parts: $(x + yi) + (c + di) = (x + c) + (y + d)i$.

Multiply $u(w+z)$

Now, multiply $u$ by the sum we obtained in Step 2, $u(w+z)$: $(a + bi)((x+c)+(y+d)i) = (a + bi)(x +c + (y+d)i)$. We can now distribute $a+bi$ to both terms in the sum: $(a+bi)(x+c) + (a+bi)(y+d)i$, which becomes $(a(x + c) - b(y + d)) + (a(y + d) + b(x + c))i$. This simplifies to: $(ax + ac - by - bd) + (ay + ad + bx + bc)i$.

Calculate $u w$ and $u z$

Next, multiply $u$ by $w$ and $z$: $u w = (a + bi)(x + yi) = (ax - by) + (ay + bx)i$, $u z = (a + bi)(c + di) = (ac - bd) + (ad + bc)i$.

Calculate the sum $u w + u z$

Now, add the results of $u w$ and $u z$: $(ax - by) + (ay + bx)i + (ac - bd) + (ad + bc)i$. This simplifies to: $(ax + ac - by - bd) + (ay + ad + bx + bc)i$.

Compare the results

Finally, we compare the expressions we obtained in Step 3 and Step 5. Both are equal: $(ax + ac - by - bd) + (ay + ad + bx + bc)i = (ax + ac - by - bd) + (ay + ad + bx + bc)i$. Thus, the given property holds true for all complex numbers $u, w, z$, and we have proved that addition and multiplication of complex numbers satisfy the distributive property: $u(w+z) = uw + uz$.

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91影视

Show that addition and multiplication of complex numbers satisfy the distributive property, meaning that $$ u(w+z)=u w+u z $$ for all complex numbers \(u, w,\) and \(z\).

Short Answer

Step by step solution

Represent the complex numbers

Calculate \(w+z\)

Multiply \(u(w+z)\)

Calculate \(u w\) and \(u z\)

Calculate the sum \(u w + u z\)

Compare the results

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