Is $2 + i > 1 - i$ true?
First off, what does it even mean to be "bigger" or "smaller" as a number?
If we say that $a>b$ means that $a$ is on the right of $b$ on the number line, could we also say that $a$ V $b$ means that $a$ is below $b$? Would this mean that we could compare "sizes" of complex numbers with an angle?
So I think that $2 + i > 1 - i$ and $2 + i$ ᴧ $1 - i$. But this compares only one part at a time, the real and then the imaginary.
So maybe $2 + i \angle_{_{angle}} 1 - i$ where $\angle_{_0}$ means $>$, $\angle_{_\pi}$ means $<$ and so on clockwise? Would that be a correct way to compare complex numbers?
Complex numbers aren't ordered in any interesting way. (Since bijections of domain $\Bbb C$ and range $\Bbb R$ exist you can order $\Bbb C$, but not in any way that satisfies useful rules.) In particular no set $P\subseteq\Bbb C$ of "positive" complex numbers exists satisfying $$0\notin P,\,\forall z\in\Bbb C\backslash\{0\}(z\in P\iff -z\notin P),\,a,\,b\in P\implies a+b,\,ab\in P,$$ because thanks to $a\in P\implies a^2\in P$ some sign $\pm i$ would satisfy $\pm i\in P\implies -1\in P\implies 1\in P$. Thus we cannot define $a>b$ as $a-b\in P$ as with real numbers viz. $P:=\Bbb R^+$.