Comparing complex numbers If $a+ib$, $c+id$, $e+if$ are three complex numbers, than can we tell which one is greater or smaller between them? If yes, then how and if no then why not? 
Can somebody give explanation on this.... I will be grateful to him. 
 A: We can define a partial order on $\Bbb C$ by $z_1\prec z_2$ if and only if $|z_1|<|z_2|$.
We can define a total order on $\Bbb C$ in various ways--I'll give you a few if you're interested.
We cannot give an order that is compatible with the operations on $\Bbb C$ so that $\Bbb C$ is an ordered field. If we could, would $i$ be positive or negative?
Basically, it depends on how you want to define "bigger/smaller" in this instance. Give us more detail, and we can better answer your question.
A: You can't define a total order $\le$ so that $\mathbb{C}$ is an ordered field:
$(1)\space\forall a, b, c \in \mathbb{C}, a ≤ b \Rightarrow a + c ≤ b + c$
$(2)\space\forall a, b \in \mathbb{C}, 0 ≤ a \land 0 ≤ b \Rightarrow 0 ≤ a b$

Either $0 \le 1$ or $1 \le 0$ in which case $0\le-1$. Let $\varepsilon \in\{-1,1\}$ so that $0 \le \varepsilon$
Either $0\le i$ or $i \le 0$ in which case $0\le-i$. Let $\delta \in \{-i,i\}$ so that $0\le\delta$
Now you can just derive something absurd:
You have $0\le \varepsilon$ and $0\le \delta$
So by $(2)$, $0\le\varepsilon\delta$
By $(2)$ again, $0\le\varepsilon\delta\delta = \varepsilon(-1) = -\varepsilon$
So $\varepsilon \le 0$
Now we have $0\le\varepsilon$ and $\varepsilon\le 0$ so $\varepsilon=0$
But we have $\varepsilon\in\{-1,1\}$
Absurd.
A: I'm thinking that you'd want to find their magnitude. Given a complex number $z = a +ib, b \ne0$, the magnitude is given by $|z| = \sqrt{z\cdot z^*} = \sqrt{(a+ib)\cdot (a-ib)} = \sqrt{a^2+b^2}$, where $z^*$ is the complex number's conjugate. Then you can order them from greatest to least in this way.
