Does anything change if we allow complex roots in system of linear equations? Say we have two variable, two equation system ($a_i, b_i, c_i \in \mathbb{R}$).
$$a_1x + b_1y + c_1 = 0$$
$$a_2x + b_2y + c_2 = 0$$
If equation $2$ is just equation $(1)$ multiplied by a constant, then the system has infinite solution.
Otherwise, if $a_1b_2=a_2b_1$, then the system has no solution.
Else, the system has a unique solution.
My question is: Does anything change if we allow $x, y$ to take complex values?
For example, is it possible for the system to have no solution for real $x, y$, but have a solution for complex $x, y$? What if the system has a unique solution, is it possible that system gets infinite solution when we extend $x, y$ to complex?
Sorry for stupid question.
Thanks
 A: Suppose we have the following system with  $a_i, b_i, c_i\in \mathbb R$: $$\begin{cases} a_1x+b_1y+c_1=0\\a_2x+b_2y+c_2=0\end{cases}\tag{*}$$
Let $x=m_1+n_1i$ and $y=m_2+n_2i$ be complex solutions to this system. This means that $$\begin{cases}(a_1m_1+b_1m_2+c_1)+(a_1n_1+b_1n_2)i=0\\ (a_2m_1+b_2m_2+c_2)+(a_2n_1+b_2n_2)i=0 \end{cases}$$ Thus, we now have two systems: $$\begin{cases}a_1m_1+b_1m_2+c_1=0\\ a_2m_1+b_2m_2+c_2=0 \end{cases}\tag{1}$$ and $$\begin{cases}a_1n_1+b_1n_2=0\\a_2n_1+b_2n_2=0 \end{cases}\tag{2}$$
Now, the second system is a homogeneous one and always has a solution $(n_1,n_2)=(0,0)$. Thus, if $(1)$ is solvable, then the equation has purely real solutions (this is also understandable from the fact that if $(1)$ is solvable, then so is $(*)$ in the reals).
But if the second system has infinite solutions, then this means that the coefficients satisfy $a_1b_2=a_2b_1$ (i.e. one equation is a multiple of the other) and the solutions are of the form $\left( t, -\dfrac{a_1}{b_1}t\right)$.
 Now, $\dfrac{a_1}{b_1}=\dfrac{a_2}{b_2}$ is already being satisfied, so in order to have a solution for $(1)$, we must have $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$, which again gives us infinite solutions, of the form $\left(z,-\dfrac{a_1z+c_1}{b_1}\right)$.
Thus, complex solutions having non-zero real and imaginary parts are possible only when one equation of $(*)$ is a multiple of the other. One may easily check that $$x=z+ti,\qquad y=-\frac{a_1z+c_1}{b_1}-\frac{a_1}{b_1}ti$$ satisfy $(*)$ for any real $z \ \text{and}\  t$.
A: Given Equations have real co-coefficients:
When given Equations have no real solution:
Let us assume that there are complex solutions but no real solution.
Then the given equation is Individually true for Complex Part and Real Part.
Then that real Part is a solution for the given equations in real numbers !
By Contradiction, there are no Complex Solutions.
[[ This is because there are no Quadratic (or higher) terms to convert Complex numbers into real numbers ]]
Given Equations have Complex co-coefficients:
It is Possible that there are no real solutions, but there may be Complex Solutions.
Eg , take $x=1+2i$ & $y=3$ , Compute 2 Independent linear combinations of $x$ & $y$ like $3x+2y$ & $x+2y$ & make your own Complex Simultaneous Equations.
That will have no real solution but will have the Complex Solution you started with.
[[ This is because we do not require Quadratic (or higher) terms to convert Complex numbers into real numbers because we allow Complex Co-efficients ]]
Quick Comment to cover Cases where real Solution Exists:
When there is one Unique real Solution, that same Solution can be taken as the Unique Complex Solution with 0 Imaginary Part. No new Solutions.
When there are infinitely many real Solutions, those same Solutions can be taken as the real Part of infinitely many Complex Solutions with Possibly Non-Zero Imaginary Part.
The Answer by user insipidintegrator has more Details on these Cases.
A: Here's yet another way to see that "nothing changes".
Recall

*

*Every complex number has a conjugate $$(a+bi)^*=a-bi$$

*Conjugates satisfy the following properties: \begin{gather*}
z^*+w^*=(z+w)^* \\
(zw)^*=z^*\cdot w^*
\end{gather*}

*$z$ is pure real iff $z=z^*$ and $z$ is pure imaginary iff $z=-z^*$.  In particular, $z+z^*$ is always purely real and $z-z^*$ is always purely imaginary.

*If $\vec{x}$ and $\vec{y}$ solve the linear system (in $\vec{x}$) $$\mathcal{A}\vec{x}=0$$ then so do $\vec{x}+\vec{y}$ and $c\vec{x}$, where $c$ is any scalar.

Extend conjugation to complex vectors by letting it act elementwise.
Suppose $\vec{f},\vec{x}\in\mathbb{C}^n$ are such that $$\mathcal{A}\vec{x}=0$$  Since that equation is just addition and multiplication, we must also have $$\mathcal{A}^*\vec{x}^*=0^*=0$$  (N.b.: $\mathcal{A}^*$ represents complex conjugation elementwise, not Hermitian conjugate.)  If the coefficients of $\mathcal{A}$ are purely real numbers, then we must have $\mathcal{A}^*=\mathcal{A}$.  That is, $\vec{x}$ solves our system iff $\vec{x}^*$ does.
In that case, not only are $\vec{x}$ and $\vec{x}^*$ solutions, but so is $\vec{x}+\vec{x}^*$, which is purely real.  Conversely, if $\vec{x}$ is a purely real solution, then (since $i$ is a scalar) $i\vec{x}$ is a purely imaginary solution.
A: If coefficients of system are real, then it’s impossible. Cramer's method to help. If some coefficients are complex, then, of course, it may be so that there is no real solution. For example, system with one equation: $$i*x=1.$$
