If $u(x,y)$ is not a harmonic function, can a harmonic conjugate $v(x,y)$ be found

Question

If $u(x,y)$ is not a harmonic function, i.e. does not satisfy Laplace equation
$$\Delta u(x,y) = \frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=0$$
(and perhaps is not continously differentiable up to degree 2), can a harmonic conjugate $v(x,y)$ be found so that
$$f(x,y)=u(x,y)+iv(x,y)$$
is analytic and therefore $v(x,y)$ is a harmonic conjugate?

Or must $u(x,y)$ be harmonic for the existence of a harmonic conjugate $v(x,y)$?

Answer 1

To answer my own question; this is my conclusion after reading and investigating.

According to Wikipedia a harmonic conjugate can be defined as follows:

A real-valued function $u(x,y)$ defined on a connected open set $\Omega \subset \mathbb R^2$ is said to have a conjugate (function) $v(x,y)$ if and only if they are respectively the real and imaginary parts of a holomorphic function $f(z)$ of the complex variable $z:=x+iy \in \Omega$

That is, $v$ is conjugate to $u$ if $f(z):=u(x,y)+iv(x,y)$ is holomorphic on $\Omega$.

A holomorphic function is defined by Wikipedia as:

A complex-valued function of one or more complex variables that is, at every point of its domain, complex differentiable in a neighborhood of the point.

It also says that "all holomorphic functions are complex analytic functions, and vice versa, is a major theorem in complex analysis".

So for $v(x,y)$ to be a conjugate to $u(x,y)$ the function $f(z)=u(x,y)+iv(x,y)$ must holomorphic or analytic.

According to Theorem 5.2 in this PDF from MIT the following is true:

$f(z)=u(x,y)+iv(x,y)$ is analytic on a region $A$ then both $u$ and $v$ are harmonic functions on $A$.

So in conclusion; since a harmonic conjugate is defined as $v(x,y)$ being the imaginary part of a holomorphic function $f(z)$ and a holomorphic function must have harmonic parts $u$ and $v$, then $u(x,y)$ must be harmonic for being able to find a harmonic conjugate.

All this is perhaps obvious since it's all in the definitions, but I wanted to post an answer regardless so it becomes obvious.