Complex numbers and their modulus

Question

Can we cancel the modulus on complex numbers?

For example: If we have $$|x + iy| = |n + im|$$ can we simply ignore the modulus on both sides? Or is that a false assumption?

Of course not. You cannot even do this for real numbers: $x=-1,y=0,n=1,m=0$ — gammatester, Jul 28 '14 at 13:59
If $r$ is a positive real, then the complex numbers $x+iy$ with modulus $r$ are all $x+iy$ such that $x^2+y^2=r^2$, so there is an entire circle of them. — André Nicolas, Jul 28 '14 at 14:02

score 0 · Answer 1 · answered Jul 28 '14 at 13:58

0

No, you can't.

Consider pairs of diametrically opposite points on the unit circle, for example.

answered Jul 28 '14 at 13:58

user_of_math

Or any pair of points on the unit circle, for that matter. – Lee Mosher Jul 28 '14 at 14:03

score 0 · Accepted Answer · answered Jul 28 '14 at 13:59

0

Of course not. $|-1|=|1|$, and $1\neq -1$

answered Jul 28 '14 at 13:59

5xum

Then the only option in solving questions with modulus on both sides is to expand them? – Gummy bears Jul 28 '14 at 13:59
@Gummybears: If you follow that option, and if you use your knowledge of Euclidean geometry and equations of circles, you will discover that it is an excellent option and that you can describe the solution set in totality. – Lee Mosher Jul 28 '14 at 14:03
Why did this vote deserve a downvote? Is there anything wrong with it? Can the downvoter please elaborate? – 5xum Jul 28 '14 at 19:06

score 0 · Answer 3 · answered Jul 28 '14 at 13:59

0

No, because two complex numbers can be different in coordinates, but have the same modulus.

Example: $z=1+i, w=1-i$.

answered Jul 28 '14 at 13:59

Mr.Fry

score 0 · Answer 4 · answered Jul 28 '14 at 14:09

0

There are an uncountably infinite number of complex numbers equal to a given valid modulus (for different values of the argument), so no.

answered Jul 28 '14 at 14:09

Wonder

4 Answers4