# bigger of two complex numbers and why?

• Jul 4th 2010, 09:47 AM
vg284
bigger of two complex numbers and why?
which one of these two complex numbers is big?
2+3i and 3+2i and why?
• Jul 4th 2010, 10:04 AM
eumyang
Define "bigger." Because I have no idea what you mean.

Are you perhaps talking about the modulus of a complex number? The modulus for a complex number a + bi is
$\displaystyle |a + bi| = \sqrt{a^2 + b^2}$.
(If you were to plot the point representing a complex number on the complex plane, the modulus is also the distance between that point and the origin.)

So plug in the numbers into the expression above to find the moduli (?) of the two complex numbers you have and you can find out which one is "bigger."
• Jul 4th 2010, 03:10 PM
mr fantastic
Quote:

Originally Posted by vg284
which one of these two complex numbers is big?
2+3i and 3+2i and why?

The complex numbers cannot be made into an ordered field so it makes no sense to ask which is bigger. The question should read

Quote:

Which one of these complex numbers has the biggest magnitude
and you have been told in post #2 how to do this.

(In fact, they both have the same magnitude so actually the question makes absolutely no sense. I suggest you point this out to your teacher).
• Jul 4th 2010, 09:10 PM
vg284

my question was

as 2<3 and 4>2 etc. Can we say that (2+3i)>(3+2i) or (2+3i)<(3+2i)?
• Jul 4th 2010, 09:20 PM
pickslides
Quote:

Originally Posted by vg284
Can we say that (2+3i)>(3+2i) or (2+3i)<(3+2i)?

You can't either of these. You can say $\displaystyle |2+3i| = |3+2i|$
• Jul 5th 2010, 04:35 AM
HallsofIvy
The set of complex numbers cannot be ordered, as uemyang and pickslides said, in such a way that the usual rules for "order" are still true:
"If a> b then a+ c> b+ c for all c"
"If a> b then ac> bc for all c> 0"
"For any a and b, one and only one must be true:
(i) a= b
(ii) a> b
(iii)b> a".

Suppose we had defined an order ">". Then, since $\displaystyle i\ne 0$, either i> 0 or i< 0.

1) Suppose i> 0. Then, multiplying both sides by the positive number i, i(i)> 0(i) or -1> 0. That by itself is NOT a contradiction since this order on the complex numbers may not give the "usual" order on the real numbers. However, now we have -1(i)> 0(i) or -i> 0. Adding i to both sides, 0> i which does contradict i> 0.

2) Suppose 0> i. Then, subtracting i from both sides 0- i> i- i or -i> 0. Multiplying both sides by the positive number -i, (-i)(-i)= i(i)> 0(-i) or -1> 0. Again, that is not necessarily a contradiction. But now, -i(-1)> -i(0) gives i> 0 which contradicts 0> i.

Either way, we arrive at a contradiction so there is no way to order the complex numbers.