<s>[INST] Problem (1) 1.A of Linear Algebra Done Right by Axler: Multiplicative Inverse My textbook, Linear Algebra Done Right, Third Edition, by Axler, has the following problem:

Suppose $a$ and $b$ are real numbers, not both $0$. Find real numbers $c$ and $d$ such that
$$\dfrac{1}{(a + bi)} = c + di.$$

This is effectively finding the multiplicative inverse $\beta = c + di$:

multiplicative inverse
for every $\alpha \in \mathbb{C}$ with $\alpha \not= 0$, there exists a unique $\beta \in \mathbb{C}$ such that $\alpha \beta = 1$.

However, I'm struggling to see how this is done.
I found the following solution online:

Because $(a + bi)(a - bi) = a^2 + b^2$, one has
$$\dfrac{1}{a + bi} = \dfrac{a - bi}{a^2 + b^2}$$
Hence
$$c = \dfrac{a}{a^2 + b^2}, \ \ d = -\dfrac{b}{a^2 + b^2}$$

However, I don't see how $c = \dfrac{a}{a^2 + b^2}, \ \ d = -\dfrac{b}{a^2 + b^2}$ follows from $\dfrac{1}{a + bi} = \dfrac{a - bi}{a^2 + b^2}$?
I would greatly appreciate it if people would please take the time to clarify this.
 [/INST] You have
$$\begin{equation}\begin{aligned}
\frac{1}{a + bi} & = \frac{a-bi}{a^2+b^2} \\
& = \frac{a}{a^2+b^2} + \frac{-bi}{a^2+b^2} \\
& = \frac{a}{a^2+b^2} + \left(\frac{-b}{a^2+b^2}\right)i
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Consider the solution of what $c$ and $d$ are in the original equation of
$$\frac{1}{(a + bi)} = c + di \tag{2}\label{eq2A}$$
Comparing the real & imaginary parts as suggested in the question comment (since $a$ and $b$ are real numbers, then $\frac{a}{a^2 + b^2}$ and $\frac{-b}{a^2 + b^2}$ are also both real values), gives the solution you're asking about, i.e., that
$$c = \frac{a}{a^2 + b^2}, \ \ d = -\frac{b}{a^2 + b^2} \tag{3}\label{eq3A}$$
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