I have no doubt that many resources about Cauchy functional equations and its relatives are available. But many properties of them have been shown in MSE posts, I will try to provide here list of links to the questions I am aware of. I have made this post CW, feel free to add more links and improve this answer in any way. Several links have been collected also in this post: Functional Equation $f(x+y)=f(x)+f(y)+f(x)f(y)$
$\newcommand{\R}{\mathbb{R}}\newcommand{\Q}{\mathbb{Q}}\newcommand{\Zobr}[3]{{#1}\colon{#2}\to{#3}}$
Cauchy's additive functional equation
We are interested in functions $\Zobr f{\R}{\R}$ such that
$$f(x+y)=f(x)+f(y) \tag{0}\label{0}$$
holds for each $x,y\in\R$.
- If $f$ is a solution of \eqref{0} and $q$ is a rational number, then $f(qx)=qf(x)$ holds for each $x\in\R$.
See e.g. If $f(x + y) = f(x) + f(y)$ showing that $f(cx) = cf(x)$ holds for rational $c$
- Every continuous solution $\Zobr f{\R}{\R}$ of \eqref{0} has the form $f(x)=cx$ for some constant $c\in\R$.
See e.g. this question: I want to show that $f(x)=x.f(1)$ where $f:R\to R$ is additive. or I need to find all functions $f:\mathbb R \rightarrow \mathbb R$ which are continuous and satisfy $f(x+y)=f(x)+f(y)$
- If $\Zobr f{\R}{\R}$ is a solution of \eqref{0} which is continuous at some point $x_0$, then it is continuous everywhere.
See e.g. Proving that an additive function $f$ is continuous if it is continuous at a single point
- If a solution $f$ of \eqref{0} is bounded on some interval, then $f(x)=cx$ for some $c\in\R$.
One part of this question is about the proof that locally bounded solution of \eqref{0} is necessarily continuous: Real Analysis Proofs: Additive Functions
- Every monotonic solution $\Zobr f{\R}{\R}$ of \eqref{0} has the form $f(x)=cx$ for some constant $c\in\R$.
See e.g. this question: Suppose $f(x)$ is linear i.e. $f(x + y) = f(x) +f(y) $ and monotone on $[-\infty, +\infty]$, then $f(x) = ax$, a real.
- If we have the equation \eqref{0} for function $\Zobr f{[0,\infty)}{[0,\infty)}$, then every solution is of the form $f(x)=cx$.
See If $f:[0,\infty)\to [0,\infty)$ and $f(x+y)=f(x)+f(y)$ then prove that $f(x)=ax$
- There exist non-continuous solutions of \eqref{0}.
See non-continuous function satisfies $f(x+y)=f(x)+f(y)$,
Do there exist functions satisfying $f(x+y)=f(x)+f(y)$ that aren't linear?,
On sort-of-linear functions, Many other solutions of the Cauchy's Functional Equation
NOTE: The proof of this fact needs at least some form of Axiom of Choice - it is not provable in ZF.
For the role of AC in this result, see this MO thread, and theses questions: Cauchy functional equation with non choice, A question concerning on the axiom of choice and Cauchy functional equation, Is there a non-trivial example of a $\mathbb Q-$endomorphism of $\mathbb R$?, What is $\operatorname{Aut}(\mathbb{R},+)$?
- The graph $G(f)=\big\{\big(x,f(x)\big)\big| x\in\R\big\}$ is a dense subset of $\R^2$ for every non-continuous solution of \eqref{0}.
See e.g. Graph of discontinuous additive function is dense in $ \mathbb R ^ 2 $
- Every anti-differentiable solution $\Zobr f{\R}{\R}$ of \eqref{0} has the form $f(x)=cx$ for some constant $c\in\R$.
See e.g. Solution of Cauchy functional equation which has an antiderivative
- Every locally integrable solution $\Zobr f{\R}{\R}$ of \eqref{0} has the form $f(x)=cx$ for some constant $c\in\R$.
See e.g. How to prove $f(x)=ax$ if $f(x+y)=f(x)+f(y)$ and $f$ is locally integrable
- Every Lebesgue measurable solution $\Zobr f{\R}{\R}$ of \eqref{0} has the form $f(x)=cx$ for some constant $c\in\R$.
See e.g. Additivity + Measurability $\implies$ Continuity or Show that $f(x+y)=f(x)+f(y)$ implies $f$ continuous $\Leftrightarrow$ $f$ measurable or Measurable Cauchy Function is Continuous or Prove that if a particular function is measurable, then its image is a rect line or Proving that $f$ is measurable with $f(x+y)= f(x)+f(y)$ then $f(x) =Ax$ for some $A\in\Bbb R$?.
Related equations
There are several functional equations that are closely related to Cauchy equation. They can be reduced to it by some methods or solved by very similar methods as the Cauchy equation.
Cauchy's exponential functional equation
$$f(x+y)=f(x)f(y) \tag{1}\label{1}$$
- If $f$ is a solution of \eqref{1}, then either $f=0$ (i.e., $f$ is constant function equal to zero) or $f(x)>0$ for each $x\in\R$.
See e.g. Is there a name for function with the exponential property $f(x+y)=f(x) \cdot f(y)$? and A non-zero function satisfying $g(x+y) = g(x)g(y)$ must be positive everywhere
If $f$ is a solution of \eqref{1}, $x\in\Q$ and if we denote $a=f(1)$, then $f(x)=a^x$.
Every continuous solution of \eqref{1} has the form $f(x)=a^x$ for some $a\ge 0$.
See e.g. continuous functions on $\mathbb R$ such that $g(x+y)=g(x)g(y)$ or How do I prove that $f(x)f(y)=f(x+y)$ implies that $f(x)=e^{cx}$, assuming f is continuous and not zero?
- If a solution of \eqref{1} is continuous at $0$, then it is continuous everywhere.
If $f\colon \mathbb{R} \to \mathbb{R}$ is such that $f (x + y) = f (x) f (y)$ and continuous at $0$, then continuous everywhere
- There are non-continuous solutions of \eqref{1}.
They can be obtained from non-continuous solutions of \eqref{1}.
See Is the group isomorphism $\exp(\alpha x)$ from the group $(\mathbb{R},+)$ to $(\mathbb{R}_{>0},\times)$ unique?
- If a solution of \eqref{1} is differentiable at $0$, then it is differentiable everywhere and $f'(x)=f(x)f'(0)$.
See Prove that $f'$ exists for all $x$ in $R$ if $f(x+y)=f(x)f(y)$ and $f'(0)$ exists and Differentiability of $f(x+y) = f(x)f(y)$ There are also posts searching for differentiable solutions: Solution for exponential function's functional equation by using a definition of derivative
Cauchy's logarithmic functional equation
$$f(xy)=f(x)+f(y) \tag{2}\label{2}$$
See Examples of functions where $f(ab)=f(a)+f(b)$ and Is the product rule for logarithms an if-and-only-if statement?
Cauchy's multiplicative functional equation
$$f(xy)=f(x)f(y) \tag{3}\label{3}$$
- Every continuous solution of \eqref{3} has the form $f(x)=x^a$
See If $f(xy)=f(x)f(y)$ then show that $f(x) = x^t$ for some t
- A solution of \eqref{3} which is continuous at $1$ is continuous for each $x>0$.
See Functional equation $f(xy)=f(x)+f(y)$ and continuity
- If a function fulfills both \eqref{0} and \eqref{3}, then either $f(x)=0$ or $f(x)=x$ (i.e., it is either zero function or the identity).
Notice that we do not require continuity here. See Functional equations $f(x+y)= f(x) + f(y)$ and $f(xy)= f(x)f(y)$
I'd rather go this way: write $\;z_1=(a,b)\;,\;\;z_2=(x,y)\;,\;\;a,b,x,y\in\Bbb R\;$ .
Then, $\;z_1\,,\;z_2\;$ are parallel iff
$$\exists\;k\in\Bbb R\;\;s.t.\;\;(x,y)=k(a,b)=(ka,kb)\iff$$
$$ z_1\overline z_2=(a,b)(ka,-kb):=\left(k(a^2+b^2)\;,\;k(-ab+ab)\right)=\left(k(a^2+b^2)\;,\;0\right)\iff$$
$$\iff\text{Im}\,(z_1\overline z_2)=0$$
Note: the last backward implication ($\Leftarrow$) is not obvious but can be deduced as follows:
$$\text{Im}\,(z_1\overline z_2)=0 \implies \text{Im}\,((ax+by,bx-ay))=0\implies bx-ay=0\implies \frac{x}{a}=\frac{y}{b}=k$$
Best Answer
What follows is just and idea.
Assume $u$ and $v$ are continuous. Let $g(x)=u^2(x)+v^2(x).$ Then $g(x+y)=g(x)g(y).$ Hence a $g(x)=e^{2\alpha x}$ for some $\alpha\in \mathbb{R}.$ For fixed $x$ there is a unique $t_x\in [0,2\pi)$ such that $$\cos(t_x)={u(x)\over \sqrt{g(x)}},\quad \sin(t_x)={v(x)\over \sqrt{g(x)}}$$ The equations involving $u(x+y)$ and $v(x+y)$ imply $$\cos(t_{x+y})=\cos(t_x+t_y),\quad \sin(t_{x+y})=\sin(t_x+t_y)$$ Hence $$t_{x+y}=t_x+t_y,\quad t_x+t_y<2\pi\\ t_{x+y}=t_x+t_y-2\pi,\quad t_x+t_y\ge 2\pi $$ Assuming continuity of $x\mapsto t_x,$ we get $t_x=\beta x.$ Hence $$u(x)=e^{\alpha x}\cos(\beta x),\qquad v(x)=e^{\alpha x}\sin(\beta x)$$