Is 8-7i > 0? where i is imaginary?

The complex field is not well-ordered.
Therefore, there is no such thing as "greater than" or less than, in complex numbers.

Let's say that the statement is true. That 8 - 7i is really greater than 0.
Then it is a positive value.

Is it bigger than 1?
If yes, then its square shoud be bigger (and positive)
(8 - 7i)^2 = (8-7i)(8-7i) = 64 - 112i - 49 = 15 - 112i

is this number (the square of 8-7i) greater than its root?

15 - 112i > 8 - 7i
subtract 8 - 7i from both sides
7 - 105i > 0

squaring it should make it even bigger:
49 - 1470i - 11,025
-10,976 - 1470i
is this number still greater than 0 AND bigger than 8 - 7i ?

I'd say no (since both components are negative)
therefore, if (8 - 7i) is positive, then it cannot be greater than 1.

If (8 - 7i) is positive and smaller than 1, then squaring it should bring it closer to 0 (and the square should remain positive). Resquaring it should bring it closer to zero still (a much smaller fraction)
In other words,
(8 - 7i)^4
the square of the square
should be very close to zero and positive.

(8 - 7i)^4 = -10,976 - 1470i
That does not sound "closer to zero" and it certainly does not look positive.

This is enough to say the statement
8-7i > 0
is "false"
----

Then, is 8-7i less than zero?
is it a negative value?

Then, the additive inverse should be positive

IF 8-7i > 0
THEN -8 + 7i > 0

redo the same attempt at a proof than we did before.

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In the end, you discover that three statements are false:

8 - 7i > 0 (is false)
8 - 7i < 0 (is false)
8 - 7i = 0 (is false)

In a well-ordered set, one (and only one) of the three must be true.

　
The concepts of greater than and less than do not exist with imaginary or complex non-real numbers.