Instructor's Solutions Manual, Section 1A Exercise 6 **6** Show that for every $\alpha \in \mathbb{C}$ with $\alpha \ne 0$ **there exists a unique** $\beta \in \mathbb{C}$ such that $\alpha \beta$ = 1. SOLUTION Suppose $\alpha = a + bi$, where $a, b \in$ $\mathbb{R}$ with at least one of $a, b$ not equal to 0. Suppose $c, d \in \mathbb{R}$ are such that $ (a + bi) (c + di) = 1.$ ( so suppose we have another complex number and it is a multiplicative inverse of $\alpha$) Multiply both sides of the equation above by $a - bi$, getting $ \begin{eqnarray} (a + bi) (a - bi) = a^2 -abi+(bi\times a) +bi\times(-bi))\\ = a^2 -b^2 (i^2)\\ = a^2 + b^2 \end{eqnarray} $ $ (a^2 + b^2) (c + di) = a - bi. $ Thus $ (a^2 + b^2)c = a, \; and \;(a^2 + b^2)d = -b, $ (by the definition of complex multiplication), which implies that $ c=\frac{a}{(a^2 + b^2)} and\; d = \frac{-b}{(a^2 + b^2)} $ The equations above show that there is at most one $\beta \in\mathbb{C}$ such that $\alpha\beta$ = 1. (Why? because these real numbers are unique. They use only $a$ and $b$, which were given ) The equations above motivate us to define $\beta \in\mathbb{C}$ , the multiplicative inverse, by $ \beta = \frac{a}{(a^2+b^2)} - \frac{b}{(a^2+b^2)}i $ The definition of complex multiplication shows that $\alpha \beta = 1.$ Try it: use den for denominator $(a^2+b^2)$ $ \begin{eqnarray} (a + bi)(\frac{a}{(a^2+b^2)} - \frac{bi}{(a^2+b^2)})\\\\ = \frac{a^2}{den} - \frac{abi}{den} + \frac{abi}{den}- \frac{b^2(i^2)}{den}\\\\ =\frac{a^2+b^2}{(a^2+b^2)} =1 \end{eqnarray} $ So we use the definition of complex multiplication to define $\beta$ And it's unique, because only uses a, b. --- Commentary: $\textbf{Exercise 6.}$ Show that for every $z\in \mathbb{C}$ with $z \neq 0$, there exists a unique $w\in \mathbb{C}$ such that $zw = 1$. $\textbf{Solution 6.}$ Suppose $z = a + bi$, where $a, b\in \mathbb{R}$ with at least one of $a, b$ not equal to $0$. Suppose $c, d\in \mathbb{R}$ are such that $(a + bi)(c + di) = 1.$ Multiply both sides of the equation above by $a - bi$, getting $(a^2 + b^2)(c + di) = a - bi.$ Thus $(a^2 + b^2)c = a \text{ and } (a^2 + b^2)d = -b,$ which implies that $c = \frac{a}{a^2 + b^2} \text{ and } d = \frac{-b}{a^2 + b^2}.$ The equations above show that there is at most one $w\in \mathbb{C}$ such that $zw = 1$. The equations above motivate us to define $w\in \mathbb{C}$ by $w = \frac{a}{a^2 + b^2} - \frac{b}{a^2 + b^2}i.$ The definition of complex multiplication shows that $zw = 1.$ $\textit{Commentary:}$ This exercise demonstrates the existence and uniqueness of multiplicative inverses (reciprocals) in the complex numbers for nonzero elements. For any nonzero complex number $z$, there is exactly one complex number $w$ such that $zw=1$. This $w$ is called the multiplicative inverse or reciprocal of $z$, and is denoted $z^{-1}$ or $\frac{1}{z}$. The proof proceeds by assuming the existence of a multiplicative inverse and deriving its form using the properties of complex multiplication. It shows that there can be at most one such inverse, establishing uniqueness. It then verifies that the derived form does indeed satisfy $zw=1$ when multiplied by $z$, confirming existence. The existence of multiplicative inverses is another crucial axiom of a field, corresponding to the division property. Together with the multiplicative identity (1 for the complex numbers), it ensures that the nonzero complex numbers form a multiplicative group. Every nonzero element has a unique inverse under multiplication, and multiplying an element by its inverse yields the identity. The formula for the multiplicative inverse, $\frac{a-bi}{a^2+b^2}$ for $z=a+bi$, has an interesting geometric interpretation. If we represent $z$ as a point $(a,b)$ in the complex plane, then $z^{-1}$ is the point obtained by reflecting $(a,b)$ across the real axis and then scaling by a factor of $\frac{1}{a^2+b^2}$. The scaling factor corresponds to the reciprocal of the squared magnitude of $z$. $\textit{Example:}$ Let $z = 3 + 4i$. Then the multiplicative inverse of $z$ is: $z^{-1} = \frac{3-4i}{3^2+4^2} = \frac{3}{25} - \frac{4}{25}i$ We can verify that $zz^{-1} = 1$: $\left(3+4i\right)\left(\frac{3}{25}-\frac{4}{25}i\right) = \frac{9}{25}+\frac{12}{25}i-\frac{12}{25}i+\frac{16}{25} = \frac{25}{25} = 1$ In $\mathbb{R}$, the multiplicative inverse of a nonzero real number $a$ is simply $\frac{1}{a}$. In $\mathbb{F}_p$ for prime $p$, the multiplicative inverse of a nonzero element $a$ is the element $b$ such that $ab \equiv 1 \pmod{p}$. For example, in $\mathbb{F}_7$, the multiplicative inverse of 3 is 5, because $3 \cdot 5 \equiv 1 \pmod{7}$. In a field of nonzero rational functions, the multiplicative inverse of a function $f(x)$ is the function $\frac{1}{f(x)}$. For nonsingular matrices, the multiplicative inverse of a matrix $A$ is the matrix $A^{-1}$ satisfying $AA^{-1}=I$. The same principle applies in other fields: the multiplicative inverse is the element that yields the multiplicative identity when multiplied with the original element. In the finite field $\mathbb{F}_p$, where $p$ is a prime number, every nonzero element has a unique multiplicative inverse. The multiplicative inverse of a nonzero element $a$ is another element $b$ in the field such that: $ab \equiv 1 \pmod{p}$ This means that when you multiply $a$ and $b$, the result is congruent to 1 modulo $p$. In your example, you mentioned $\mathbb{F}_7$, the finite field with 7 elements. To find the multiplicative inverse of 3 in this field, we need to find an element $b$ such that: $3 \cdot b \equiv 1 \pmod{7}$ By testing the elements of the field, we find that 5 satisfies this condition: $3 \cdot 5 = 15 \equiv 1 \pmod{7}$ Therefore, the multiplicative inverse of 3 in $\mathbb{F}_7$ is 5. It's important to note that the multiplicative inverse exists only for nonzero elements in the field. The element 0 does not have a multiplicative inverse because there is no element $b$ such that: $0 \cdot b \equiv 1 \pmod{p}$ This is true for any prime $p$.