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I was wondering if elements that are imaginary pure where comparable. I mean, does $3i>0$ makes sense ? And $3i<4i$ ?

May be for the first one I should write $3i>0i$ ?

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    It depends how you define $<$ in $\Bbb C$. But it is known that $\Bbb C$ is not an ordered field, unlike $\Bbb R$, this means that any order relation $<$ that you define in $\Bbb C$ it will be not compatible with the field operations. – 2017-02-25

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We cannot define an order relation in $\mathbb{C}$ compatible with the operations of addition and product and the order of $\mathbb{R}$. In case that we can do it, $i$ (which is different of zero) can't be greater than zero, because in this case $i\cdot i=-1$ must be greater than zero and it isn't. But $i$ can't be lower than zero, because in this case $-i$ would be greater than zero and then $(−i)(−i) = −1$ must be greater than zero and it isn't.

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The complex numbers are not linearly ordered in a way that's compatible with their arithmetic properties. You could define a linear ordering on pure imaginary numbers, but for what purpose?

Such an ordering could be compatible with addition, but pure imaginary numbers don't even form a closed set under multiplication, so there's not much to work with there.

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    But $i\mathbb R$ works as $\mathbb R$ no ? So 3i<4i should make sense, no ? – 2017-02-25
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    See the answer by @mannav. That could be compatible with addition, but not with multiplication. We like it when ordering relations are compatible with multiplication. – 2017-02-25