Complex numbers in System of Linear Equations Good evening,
Summing up the problem I have is the following:
Is it possible to have complex numbers in a System of Linear Equations?
It is hard for me to visualize this problem as well, since if we have a System of Linear Equations with, for example, three unknowns and three equations, it is possible to find the relationship between those three equations of lines. Visualizing this is doable as well.
But, trying to do this with complex numbers is confusing me. I do think it is possible but an approach is not coming to mind.
Kind regards,
Tema.
 A: Indeed, this is possible but you should not expect to get a nice visualisation.
First of all, all the arithmetic operations that you need to solve linear equations over the reals also work for complex numbers. So if you know the arithmetical rules for complex numbers then you can solve such a system of linear equations exactly in the same way as for real numbers.
But why is it harder to visualize this compared to a system over real numbers? Assume that you have a system of linear equations over the complex number with $3$ variables and $3$ equations. Then every variable can have a real and imaginary part now. Thus you can translate this into an equivalent system with $6$ variables and $6$ equations now. Let me give an example.
Imagine you have the following system of linear equations over the complex numbers:
$$
z + iu = 1, \\ z - u = i
$$
We can transform this into an equivalent system over the reals by writing $z = \alpha_z + i\beta_z$ and $u = \alpha_u + i\beta_u$. Plugging this into the equations yields:
$$
\alpha_z + i\beta_z + i(\alpha_u + i\beta_u) = 1, \\ \alpha_z + i\beta_z - \alpha_u + i\beta_u = i
$$
Now we can group this a bit differently:
$$
\alpha_z + - \beta_u + i(\beta_z + \alpha_u) = 1, \\ \alpha_z - \alpha_u + i(\beta_z + \beta_u) = i
$$
Now each of these equations is actually two equations (because we have a constraint on both the imaginary and the real part in each equation). So we can write this equivalently as:
$$
\alpha_z + - \beta_u = 1,\\ i(\beta_z + \alpha_u) = 0, \\ \alpha_z - \alpha_u = 0, \\i(\beta_z + \beta_u) = i
$$
As a last step we observe that we can divide the equations for the imaginary parts by $i$ in order to get a system of linear equations over the reals:
$$
\alpha_z + - \beta_u = 1,\\ \beta_z + \alpha_u = 0, \\ \alpha_z - \alpha_u = 0, \\\beta_z + \beta_u = 1
$$
This last system of equations (over the reals) is equivalent to what we started from, but it has now four variables and four constraints whereas our initial system had two variables and two constraints.
Note that you now can visualize this system of real equations (imagine $4$ lines in the plane). But it is not clear (at least to me) how this would translate to a visualization of the behavior of the complex variables $z$ and $u$.
A: Yes, you can have sets of complex linear equations. For instance,
$$
\cases{(5+3i)x+(2-i)y=1\\(-7+4i)x+(8+9i)y=4-3i}
$$
Yes, it's difficult to visualise. This one in particular requires you to see in your mind's eye two planes in four-dimensional space, and find their intersession point. And it doesn't get easier for higher number of unknowns.
However, the algebra doesn't care. You solve this more or less exactly the same way you would solve a set of two real equations in two variables. Get rid of one variable through either substitution, or adding some suitable multiple of one equation to the other equation, then solve for the other variable.
