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I have heard that you can't compare complex numbers... Well it is true because it's not defined on the real number line, you can't compare imaginary quantities... But I just wanted to know that why can't we compare them using their modulus?

For example, let's say we have two complex numbers:

$$z_1 = 4 + 3i \quad \text{ and } \quad z_2 = 7 + 9i$$

Now we take their modulus,

$$|z_1| = 16 + 9 = 25 \\ |z_2| = 49 + 81 = 130$$

Now we could say that $z_2$ is greater than the first one right?

VIVID
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Param Varsha
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    There are orders on $\mathbb C$, but they are not compatible with multiplication; cf. this question – J. W. Tanner Oct 24 '21 at 04:27
  • Right, $\mathbb C$ cannot be made an ordered field. Your order is called a “pre-order,” because $|a|\leq |b|$ and $|b|\leq |a|$ does not imply $a=b.$ – Thomas Andrews Oct 24 '21 at 04:32
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    You totally can compare their moduli, as you've demonstrated. But this comparison only claims that one has a bigger modulus, not that one is bigger than the other. If both number have the same modulus and one has a bigger principal argument, then which is the "bigger" number? – ryang Oct 24 '21 at 04:34
  • Also see https://math.stackexchange.com/questions/487997/total-ordering-on-complex-numbers – Ross Millikan Oct 24 '21 at 05:00
  • An "order" on $,\mathbb C,$ where all non-zero numbers are greater than $,0,$ is neither consistent with the usual order on $,\mathbb R,$ nor too useful. For example, $,1,$ and $,-1,$ would both be greater than $,0,$. – dxiv Oct 24 '21 at 06:22

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