Complex numbers as an abelian group

[karnten07]

[SOLVED] Complex numbers as an abelian group

Homework Statement

Multiplication of complex numbers defines a binary operation on C^x:=C\{0} (complex numbers not including zero). Show that C^x together with this binary operation is an abelian group. (without further discussion you may use the usual laws of algebra for R,
such as associativity for addition and multiplication of real numbers)

p.s the wording may sound strange because my lecturer is european i think and english is his second language.

Homework Equations

The Attempt at a Solution

I know that for an abelian it must show commutativity, ie, x*y=y*x. For it to be a group there must be associativity of multiplication, that en element exists e, that is an identity element ie. e*x=x and x*e=x, and also that there is an element y, that is an inverse element ie. x*y=e and y*x=e

I have wrote this information out in my answer but i think i might need to show it in an example. What kind of example would someone suggest, maybe (1+i) x (1-i). Giving 1 +1=2?? Any ideas guys? Thanks

 

[Dick]

Examples don't show anything. You have the properties you want to prove. Now start showing them one by one. What's the identity, e? Can you show commutativity, (a+bi)*(c+di)=(c+di)*(a+bi)? I would next show there is an inverse, and then use that to prove closure by showing a*b=0 implies a*e=0. Associativity is more of a pain in the neck to write down than anything deep. Showing you have an inverse is the only proof with any real meat on the bones.

 

[karnten07]

Examples don't show anything. You have the properties you want to prove. Now start showing them one by one. What's the identity, e? Can you show commutativity, (a+bi)*(c+di)=(c+di)*(a+bi)? I would next show there is an inverse, and then use that to prove closure by showing a*b=0 implies a*e=0. Associativity is more of a pain in the neck to write down than anything deep. Showing you have an inverse is the only proof with any real meat on the bones.

I have shown that commutativity exists, but how do i show there is an inverse, can you suggest how i compute the inverse of a complex number?

 

[Dick]

Good start. If you think the identity is 1+0i, you're right. To get the inverse of (a+bi) which is 1/(a+bi) multiply 1/(a+bi) by (a-bi)/(a-bi)=1. You are new to complex numbers, aren't you?

 

[karnten07]

Good start. If you think the identity is 1+0i, you're right. To get the inverse of (a+bi) which is 1/(a+bi) multiply 1/(a+bi) by (a-bi)/(a-bi)=1. You are new to complex numbers, aren't you?

Well i have worked with complex numbers before but not in great depth really. Oh i see, so the inverse is a-bi/(a^2 +b^2)?

 

[Dick]

Yes, so if a+bi is not 0+0i, you have a multiplicative inverse.

 

[karnten07]

Yes, so if a+bi is not 0+0i, you have a multiplicative inverse.

Brilliant, I've shown the group is abelian now pretty much, thanks again.