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mathstudent79
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Homework Statement
Hi. First post here. I searched for this problem and nothing I saw helped. Sorry. Thanks in advance for help.
define a function f: C -----> S by:
f(a,b) = a+bi where a,b [tex]\in[/tex]R
We leave it to reader to check that f is one-one and onto and that:
f((a,b)+(c,d)) = f((a,b)) + f((c,d))
and
f((a,b)(c,d))=f(a,b)f(c,d)
Homework Equations
a+bi = (a,b)
(a+bi)+(c+di) = (a+c) + (b+d)i
(a+bi)(c+di)=(ac-bd)+(bc+ad)i
The Attempt at a Solution
Here's what I've done so far.
f((a,b)+(c,d))=(a+c)+(b+d)i=(a+bi)+(c+di)=f(a,b)+f(c,d)
f((a,b)(c,d))=(ac-bd)+(bc+ad)i=(a+bi)(c+di)=f(a,b)f(c,d)
sp' f(a,b)=f(a',b')
then a+bi=a'+b'i
then (a,b)=(a',b')
but this is only true if a=a' and b=b' So f(a,b)=f(a',b') implies (a,b)=(a',b')
HERE'S where I'm especially not sure:
onto:
suppose we have a+bi. By definition of function, there is an a,b s.t. f(a,b)=a+bi. Therefore onto.
Thanks a lot guys.