There seems to be this conception that inequalities can't be applied to complex numbers. However, I don't think this is true. It is true that $<$ and $>$ cannot be used between any 2 complex numbers, but we can invent new inequality signs. First, let's focus on $<$. If $a < b$, then $b - a > 0$, and thus, $b - a$ is in the set of positive real numbers. Conversely, $a - b < 0$, and thus, $a - b$ is in the set of negative real numbers. Now, we all know that the argument of a complex number is the angle between the line that that number falls on and the positive real line. We also know that the argument of a positive real number is just $0$. Thus, if $a < b$, $\arg(b - a) = 0$.
Now we can invent a set of new inequality signs. Let us define $<_q$ such that $-p \leq_q p$. For any 2 complex numbers $a$ and $b$, $a <_q b$ implies that $\arg(b - a) = q$. We can see that $<_0$ has the same meaning as $<$. It's important to note that $<$ can be used between non-real numbers, as long as the imaginary parts are the same. If we real numbers $x$, $y$, and $z$ where $x < y$, then $x + zi < y + zi$, since $(y + zi) - (x + zi) = y - x > 0$. Another way to think of this as the if $a <_q b$, then to get from $a$ to $b$, you have to travel parallel to the line whose argument is $q$.
So what do you think? Is this a good way to define inequalities between complex numbers?