Instructor's Solutions Manual, Section 1A Exercise 5 1. 5  Show that for every 𝛼∈𝐂, there exists a unique 𝛽∈𝐂 such that 𝛼+𝛽 = 0. 5 Show that for every $\alpha \;\in\; \mathbb{C}$, there exists a unique $\beta\;\in\mathbb{C}$ such that $\alpha + \beta = 0.$ SOLUTION Suppose $\alpha = a + bi$, where $a, b\in\mathbb{R}$ . Let $\beta = -a - bi.$ Then the definition of complex addition shows that $ \alpha + \beta = 0. $ Suppose $\lambda \in \mathbb{C}$ is such that $\alpha + \lambda = 0$. Adding $\beta$ to both sides of the equation above shows that $\lambda = \beta$. $\begin{eqnarray} \alpha + \beta + \lambda = 0 + \beta\\ 0 + \lambda = \beta\\ \lambda = \beta\\. \end{eqnarray} $ Thus $\alpha$ has a unique additive inverse. --- Commentary: $\textbf{Exercise 5.}$ Show that for every $z\in \mathbb{C}$, there exists a unique $w\in \mathbb{C}$ such that $z + w = 0$. $\textbf{Solution 5.}$ Suppose $z = a + bi$, where $a, b\in \mathbb{R}$. Let $w = -a - bi$. Then the definition of complex addition shows that $z + w = 0.$ Suppose $u\in \mathbb{C}$ is such that $z + u = 0.$ Adding $w$ to both sides of the equation above shows that $u = w$. Thus $z$ has a unique additive inverse. $\textit{Commentary:}$ This exercise proves the existence and uniqueness of additive inverses in the complex numbers. For any complex number $z$, there is exactly one complex number $w$ such that $z+w=0$. This $w$ is called the additive inverse or negative of $z$, and is denoted $-z$. The proof constructs the additive inverse explicitly by negating the real and imaginary parts of $z$. It then shows that this is the only possible additive inverse, establishing uniqueness. The existence of additive inverses is another key axiom of a field. Together with the existence of an additive identity (0 for the complex numbers), it ensures that the complex numbers form an additive group. Every element has an inverse under addition, and adding an element to its inverse yields the identity. $\textit{Example:}$ Let $z = 3 - 4i$. Then the additive inverse of $z$ is $-z = -3 + 4i$, because: $z + (-z) = (3 - 4i) + (-3 + 4i) = 0 + 0i = 0$ In $\mathbb{R}^2$, the additive inverse of a vector $(a,b)$ is the vector $(-a,-b)$. Geometrically, this corresponds to the vector pointing in the opposite direction with the same magnitude. In $\mathbb{F}_p$, the additive inverse of an element $a$ is the element $b$ such that $a + b \equiv 0 \pmod{p}$. For example, in $\mathbb{F}_7$, the additive inverse of 3 is 4, because $3 + 4 \equiv 0 \pmod{7}$. In a field of functions, the additive inverse of a function $f(x)$ is the function $-f(x)$. For matrices, the additive inverse of a matrix $A$ is the matrix $-A$ obtained by negating each entry. The same principle applies in other fields: the additive inverse is obtained by applying the field's negation operation to the element.