- #1
dancergirlie
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Homework Statement
Show that the complex conjugation function f:C----->C (whose rule is f(a+bi)=a-bi) is a bijection
Homework Equations
A function is a bijection if it is both injective and surjective
a function is injective if when f(a)=f(b) then a=b
a function is surjective if for f:B---->C
if c is in C then there exists a b in B such that f(b)=c
The Attempt at a Solution
I get the whole injective part
Let
f(a+bi)=f(c+di)
hence, a-bi=c-di
now suppose that b is unequal to d
so, a-c=(b-d)i
meaning i=(a-c)/(b-d) which is a contradiction, since i is not a real number.
hence, a=c and b=d
meaning, f is injective
My problem is with the surjective part, let me know if this is wrong,
Let f(a+bi)=c+di
then, a-bi=c+di... then i don't know where to go from there
I don't know if this is wrong, any clarification on how to test for surjectivity would be greatly appreciated...